The probability that a device using five silicon chips selected randomly from a batch of 100 chips, which includes five defective ones, contains no defective chip is calculated by the ratio of selecting five good chips to selecting any five chips from the batch.
Explanation:The question is asking to find the probability that a device, which uses five silicon chips selected from a batch of a hundred chips with five being defective, will have no defective chip. To solve this, we can calculate the probability step by step using the concept of combinations.
Firstly, we determine the number of ways to select five non-defective chips out of 95 good ones, which is C(95,5). Then, we calculate the total number of ways to select any five chips out of the whole batch, which is C(100,5). The probability that the device contains no defective chip is the ratio of these two numbers:
P(device has no defective chip) = C(95,5) / C(100,5)
Where C(n,k) represents the number of combinations of n items taken k at a time.
To calculate this, use factorials where C(n,k) = n! / [(n-k)!k!].
So, the probability that the device contains no defective chip, is:
P(device has no defective chip) = (95! / (90!*5!)) / (100! / (95!*5!))
Simplifying the factorials, we have:
P(device has no defective chip) = (95*94*93*92*91) / (100*99*98*97*96)
Finally, calculate this to get the decimal value, which would give the probability that the device contains no defective chips when made up from one batch.
Consider the following function.
f(x) = (4 − x)e−x
(a) Find the intervals of increase or decrease. (Enter your answers using interval notation.)
increasing
decreasing
(b) Find the intervals of concavity. (Enter your answers using interval notation. If an answer does not exist, enter DNE.)
concave up
concave down
(c) Find the point of inflection. (If an answer does not exist, enter DNE.)
(x, y) =
Answer:
a) decreases at interval (-∞,5) and increases at (5,∞)
b) is convave down at interval (-∞,6) an up at interval (6,∞)
c) f(x) has an inflexion point at x=6
Step-by-step explanation:
a) for the function
f(x) = (4 − x)*e^(−x)
then the derivative of f(x) indicates if the function decreases or increases. Thus
f'(x) =df(x)/dx = -e^(−x) -(4 − x)*e^(−x)= (x-5)*e^(−x)
since e^(−x) is always positive , then
f'(x) < 0 for x<5 → f(x) decreases when x<5 ( interval (-∞,5) )
f'(x) > 0 for x>5 → f(x) increases when x>5 ( interval (5,∞) )
f'(x) = 0 for x=5 → f(x) has a local minimum ( since first decreases and then increases)
b) the concavity is found with the second derivative of f(x) , then
f''(x) =d²f(x)/(dx)² = e^(−x) - (x-5)*e^(−x) = (6-x)*e^(−x)
then
f''(x) < 0 for x>6 → f(x) is convave up for x>6 ( interval (6,∞) )
f'(x) > 0 for x<6 → f(x) is concave down when x<6 ( interval (-∞,6) )
f'(x) = 0 for x=6 → f(x) has an inflection point at x=6
The function f(x) = (4 - x)e^-x is decreasing on the interval (-∞, 1), increasing on (1, ∞), concave up on (2, ∞), concave down on (-∞, 2), and has its point of inflection at [tex](2, 2e^-^2).[/tex]
Explanation:To find the intervals of increase or decrease, we first need to find the derivative f'(x) of the function f(x) = (4 - x)e-x. Using the product rule and chain rule, we find [tex]f'(x) = e-x - (x - 4)e-x.[/tex]
Setting f'(x) to zero and solving for x, we find the critical points x = 1 and x = 4. Using test points, we determine that the function is decreasing on (-∞, 1) and increasing on (1, ∞).
Next, we find the second derivative [tex]f''(x) = -2e-x + 2(x - 4)e-x[/tex] to determine concavity. Setting f''(x) equal to zero and solving for x, we find x = 2. Using test points, we find that the function is concave up on (2, ∞) and concave down on (-∞, 2). Since the function changes concavity at x = 2, this is the point of inflection.
Substituting x = 2 into the original function f(x), we find y = 2e-2, so the point of inflection is (2, 2e-2).
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Find the probability that the age of a randomly chosen American (a) is less than 20. (b) is between 20 and 49. (c) is greater than 49. (d) is greater than 29
Answer: i think its B
Step-by-step explanation:
The following probability distributions of jobsatisfaction scores for a sample of informationsystems (IS) senior executives and IS middle managersrange from a low of 1 (very dissatisfied) to a high of5 (very satisfied).Probability Job Satisfaction Score IS SeniorExecutives 1 .05 2 .093 .03 4 .425 .41IS Middle Managers.04.10.12.46.28a. What is the expected value of the job satisfactionscore for senior executives?b. What is the expected value of the job satisfactionscore for middle managers?c. Compute the variance of job satisfaction scores forexecutives and middle managers.d. Compute the standard deviation of job satisfactionscores for both probability distributions.e. Compare the overall job satisfaction of seniorexecutives and middle managers.
Answer:
a) 4.076
b) 3.9
c) variance for executives=1.128
variance for middle mangers=0.73
d)standard deviation for executives=1.062
standard deviation for middle mangers=0.854
e) Overall job satisfaction for senior executives is higher than middle manager.
Step-by-step explanation:
IS senior executives
Job Satisfaction 1 2 3 4 5
Probability 0.05 0.093 0.03 0.425 0.41
IS middle manager
Job Satisfaction 1 2 3 4 5
Probability 0.01 0.1 0.12 0.46 0.28
Let X denotes IS senior executive and Y denotes IS middle manager.
a)
E(X)=∑x*p(x)=1*0.05+2*0.093+3*0.03+4*0.425+5*0.41
E(X)=0.05+0.186+0.09+1.7+2.05
E(X)=4.076
b)
E(Y)=∑y*p(y)=1*0.1+2*0.1+3*0.12+4*0.46+5*0.28
E(Y)=0.1+0.2+0.36+1.84+1.4
E(Y)=3.9
c)
V(x)=∑x²*p(x)-(∑x*p(x))²
∑x²*p(x)=1*0.05+4*0.093+9*0.03+16*0.425+25*0.41
∑x²*p(x)=0.05+0.372+0.27+6.8+10.25
∑x²*p(x)=17.742
V(x)=17.742-(4.076)²
V(x)=1.128
V(y)=∑y²*p(y)-(∑y*p(y))²
∑y²*p(y)=1*0.1+4*0.1+9*0.12+16*0.46+25*0.28
∑y²*p(y)=0.1+0.4+1.08+7.36+7
∑y²*p(y)=15.94
V(y)=15.94-(3.9)²
V(y)=0.73
d)
S.D(x)=√V(x)
S.D(x)=√1.128
S.D(x)=1.062
S.D(y)=√V(y)
S.D(y)=0.854
e)
Overall job satisfaction for senior executives is more than middle manager as expected value of senior executives is greater than expected value of middle manger with relatively higher variability than middle manager.
a 50m long chain hangs vertically from a cunlinder attached to a winch. Assume there is no friction in the system and that the chain has a density of 10kg/m. how much work is required to wind the chain into the cylinder if a 50kg block is attached to the end of the chain?
Answer:
147000 J
Step-by-step explanation:
We are given that
Length of chain=L=50 m
Density of chain=[tex]\rho=10kg/m^3[/tex]
We have to find the work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain.
Work done=[tex]\int_{a}^{b}F(y)dy[/tex]
We have F(y)=[tex]\rho g(50-y)dy[/tex]
a=0 and b=50
[tex]g=9.8m/s^2[/tex]
Using the formula
Work done=[tex]w_1=10\times 9.8\int_{0}^{50}(50-y)dy[/tex]
Where Length of chain is (50-y) has to be lifted.
Work done=[tex]w_1=10\times 9.8[50y-\frac{y^2}{2}]^{50}_{0}[/tex]
By using the formula [tex]\int x^ndx=\frac{x^{n+1}}{n+1}+C[/tex]
Work done=[tex]w_1=10\times 9.8\times (50(50)-\frac{(50)^2}{2})=98\times (2500-1250)=122500 J[/tex]
When the chain is weightless then the work done required to lift the block attached to the 50 m long chain
Again using the formula
Where f(y)=mg
[tex]w_2=\int_{0}^{50}mgdy[/tex]
We have m=50 kg
[tex]w_2=\int_{0}^{50}50\times 9.8 dy=490[y]^{50}_{0}=490\times 50=24500 J[/tex]
The work done required to wind the chain into the cylinder if a 50 kg block is attached to the end of the chain=[tex]w_1+w_2=122500+24500=147000 J[/tex]
A force of 10 lb is required to hold a spring stretched 2 in. beyond its natural length. How much work W is done in stretching it from its natural length to 5 in. beyond its natural length?
Answer:
Work done will be equal to 5.2059 lb-ft
Step-by-step explanation:
We have given force F = 10 lb
Spring is stretched to 2 in
So x = 2 in
As 1 inch = 0.0833 feet
So 2 inch = 2×0.0833 = 0.1666 feet
From hook's law we know that F = Kx , here K is spring constant and x is spring elongation
So [tex]10=K\times 0.1666[/tex]
K = 60.024 lb/feet
Now new elongation x = 5 in
So 5 in = 5×0.0833 = 0.4165 feet
Work done is given by [tex]W=\frac{1}{2}Kx^2[/tex]
So [tex]W=\frac{1}{2}\times 60.02\times 0.4165^2=5.205lb-ft[/tex]
So work done will be equal to 5.2059 lb-ft
In a box there are two coins: a standard coin with head and tail and a 2-headed coin. You randomly pick one of the coins, toss it and see a head. What is the probability that the other side of this coin is a head?
Answer:
Probability that the other side of this coin is a head = 0.5
Step-by-step explanation:
Given that there are two coins: a standard coin with head and tail and a 2 - headed coin.
When tossing a randomly chosen coin once the sample space obtained will be :
Head of a standard coin.Tail of a standard coin.Front Head side of 2-headed coin.Back Head side of 2-headed coin.Now since we have to find the probability that the other side of this coin which is tossed is also a head which means probability of selecting a 2-headed coin.
From the above cases there are two outcomes which are in favour from total of four outcomes;
Hence, Probability that the other side of this coin is a head = [tex]\frac{1}{2}[/tex] = 0.5 .
Waiting in line. A quality - control manager at an amusement park feels that the amount of time that people spend waiting in line for the American Eagle roller coaster is too long. To determinate if a new loading/unloading procedure if efective in reducing wait time in line, he measured the amount of time (in minutes) people are waiting in line for 7 days. After implementing the new procedure, he again measures the amount of time in minutes and people are waiting in line 7 days and obtains the following data.
Wait time before new procedure
Day
Mon Tues Wed Thurs Fri Sat Sat Sun Sun
11.6 25.9 20.0 38.2 57.3 32.1 81.8 57.1 62.8
Wait time after new procedure
10.7 28.3 19.2 35.9 59.2 31.8 75.3 54.9 62.0
test the claim that the new loading/unloading procedure is effective in reducing wait time (H0: µd=0 and H1: µd<0)at α=.05 level of significance. Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers (use the classical approach and the p-value approach).
Answer:
No
explanation:
given:
n=9
[tex]\alpha[/tex]=0.05
see the attachment
Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.
d=0.9-2.4+0.8+...+6.5+2.2+0.8/9
=1.0556
The variance is the sum of squared deviations from the mean divided by n-1. The standard deviation is the square root of the variance. Determine the sample standard deviation of the differences:
s_d=√(0.9-1.0556)^2+...+(0.8-1.0556)^2/9-1
=2.6
CLASSICAL APPROACH :
Given claim: new procedure reduces [tex]u_{d}[/tex] > 0
The claim is either the null hypothesis or the alternative hypothesis The null hypothesis and the alternative hypothesis state the opposite of each other The null hypothesis needs to contain an equality
[tex]H_{0}:u_{d}=0\\ H_{1}:u_{d}>0[/tex]
Determine the value of the test statistic
t=d-[tex]u_{d}[/tex]/s_d/√n
=1.220
Determine the critical value from the Student T distribution table in the appendix in the row with d_f = n- 1 = 9-1 = 8 and in the column with [tex]\alpha[/tex] = 0.05 t =1.860
The rejection region then contains all values larger than 1.860
If the value of the test statistic is within the failed region, then the null hypothesis is failed
1.220 < 1.860 failed H_0
There is not sufficient evidence to support the claim that the new loading/unloading procedure is effective in reducing the wait time.
P VALUE APPROACH:
Given claim: new procedure reduces [tex]u_{d}[/tex] > 0
The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain an equality.
[tex]H_{0}:u_{d}=0\\ H_{1}:u_{d}>0[/tex]
Determine the value of the test statistic:
t=d-[tex]u_{d}[/tex]/s_d/√n
The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Students T distribution in the appendix containing the t-value in the row d_f = n-1 = 9-1 = 8
0.10 < P < 0.15
If the P-value is less than the significance level, reject the null hypothesis.
P > 0.05 failed H_0
There is not sufficient evidence to support the claim that the new loading unloading procedure is effective in reducing the wait time.
dentify the type of data (qualitative/quantitative) and the level of measurement for the native language of survey respondents. Explain your choice. Native language Number of respondents English 759 Spanish 775 French 22 Are the data qualitative or quantitative? A. Quantitative, because descriptive terms are used to measure or classify the data. B. Qualitative, because descriptive terms are used to measure or classify the data. C. Qualitative, because numerical values, found by either measuring or counting, are used to describe the data. D. Quantitative, because numerical values, found by either measuring or counting, are used to describe the data.
Answer:
The correct option is D i.e. Quantitative because numerical values found by either measuring or counting are used to describe the data.
Step-by-step explanation:
As the number of respondents is a numerical value and is identified by counting thus it is a quantitative variable. Also all the other options are incorrect.
A is incorrect because the reason described is not the property of quantitative data.
B is incorrect because the data is not described in descriptive terms.
C is incorrect because the reason described in not a property of qualitative data.
I need help please!!!!
Answer:
x = 14.48
Step-by-step explanation:
first we have to see that we have the measurements from all sides
and we know that the angle between side 21 and 20 is 90 degrees
well to start we have to know the relationships between angles, legs and the hypotenuse.
a: adjacent
o: opposite
h: hypotenuse
sin α = o/h
cos α= a/h
tan α = o/a
let's take the left angle as α
sin α = 21/29
α = sin^-1 (21/29)
α = sin^-1 (0.7241)
α = 46.397
Now we do the same with the smaller triangle
tan α = o/a
sin 46.397 = x/20
0.724 = x/20
0.724 * 20 = x
14.48 = x
x = 14.48
if we want to check it we can do the same procedure with the other angle
If a distribution has "fat tails," it exhibits A. positive skewness B. negative skewness C. a kurtosis of zero. D. excess kurtosis. E. positive skewness and kurtosis.
Answer: D. Excess Kurtosis
Step-by-step explanation:
A fat tailed distribution is a kind of probability distribution that exhibits excess kurtosis because it means the resulting numbers from the probability distribution are on a large scale power increment or very small/ slow decreeing order. This makes the graph on the distribution literally fat tailed and makes skewness in such distribution data extremely difficult to ascertain.
Fat tails in a distribution signify excess kurtosis, which signifies more extreme values or outliers than in a normal distribution. Neither positive skewness, negative skewness, nor a kurtosis of zero signify a distribution's 'fat tails'.
Explanation:If a distribution has 'fat tails', it represents 'excess kurtosis'. This term is used to describe a distribution of data that features tails that are fatter and longer than in a normal distribution. This often means the distribution exhibits more extreme values or outliers. When a distribution has excess kurtosis, it has strong outliers.
Positive skewness, negative skewness, and a kurtosis of zero have no correlation with 'fat tails'. While skewness refers to the asymmetry of a distribution, and a kurtosis of zero refers to a normal distribution, neither of these refer to the concept of 'fat tails'.
So, Fat tails in a distribution signify excess kurtosis, not positive skewness, negative skewness, or a kurtosis of zero.
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For each gym class a school has 10 soccer balls and 6 volleyballs all of the classes share 15 basketballs. The expression 10c+6c+15 represents the total number of balls the school has for c classes what is a simpler form of the expression
Answer:
[tex]16c+15[/tex]
Step-by-step explanation:
we have the expression
[tex]10c+6c+15[/tex]
step 1
We can simplify the expression by combining like terms. That is, the terms with the same variable
[tex](10c+6c)+15[/tex]
[tex]16c+15[/tex]
Answer: 16c+15
Step-by-step explanation:
Step 1
Write down the given expression (10c+6c)+15
Step 2
10+6 = 16c
So, 16c+15
Hope this helps! (✿◡‿◡)
A study is designed to test the effect of light level on exam performance of students. The researcher believes that light levels might have different effects on males and females, so wants to make sure both arc equally represented in each treatment. The treatments are fluorescent overhead lighting, yellow overhead lighting, no overhead lighting (only desk lamps).
(a) What is the response variable?
(b) What is the explanatory variable? What arc its levels?
(c) What is the blocking variable? What arc its levels?
In the given study the response variable is the exam performance, the explanatory variable is the light level, and the blocking variable is the gender of the students.
Explanation:In this study, the response variable is the exam performance of the students. This is what is being measured as an outcome. The explanatory variable is the light level. The different light levels (fluorescent overhead lighting, yellow overhead lighting, or no overhead light and only desk lamps) constitute the treatments are its levels. The blocking variable in this case is the gender of the students. By making sure that both males and females are equally represented in each treatment, the researcher is controlling for the effect of gender. The levels of this blocking variable are male and female.
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The study measures the effect of light level (explanatory variable) on student exam performance (response variable), with gender as a blocking variable. Lurking variables and study design elements like random assignment and blinding are crucial for maintaining the validity of the results.
a) In the study described, the response variable is the exam performance of the students. This variable will be measured to assess the impact of different lighting conditions on students' ability to perform on an exam.
b) The explanatory variable, or independent variable, is the type of lighting. The levels of this variable are fluorescent overhead lighting, yellow overhead lighting, and no overhead lighting (only desk lamps).
c) The blocking variable is gender. The researcher wants to make sure that both males and females are equally represented in each treatment to test the hypothesis that light levels might affect genders differently. The levels of this variable are simply male and female.
When selecting participants, it is important to consider random assignment to ensure that each treatment group is similar in all respects other than the treatment itself. The idea of dividing participants to drive without distraction and to text and drive could be problematic due to ethical considerations and the introduction of confounding variables.
Lurking variables that could interfere with the study on light levels might include the time of day the exam is taken, students' prior knowledge and preparation levels, or even the difficulty of the exam itself.
Blinding could be used by ensuring that the person measuring exam performance does not know which lighting condition the student was exposed to, thus preventing any bias in the evaluation of the exam performance.
Matt is a software engineer writing a script involving 6 tasks. Each must be done one after the other. Let ti be the time for the ith task. These times have a certain structure:
•Any 3 adjacent tasks will take half as long as the next two tasks.
•The second task takes 1 second.
•The fourth task takes 10 seconds.
a) Write an augmented matrix for the system of equations describing the length of each task.
b) Reduce this augmented matrix to reduced echelon form.
c) Suppose he knows additionally that the sixth task takes 20 seconds and the first three tasks will run in 50 seconds. Write the extra rows that you would add to your answer in b) to take account of this new information.
d) Solve the system of equations in c).
Answer:
Let [tex]t_i[/tex] be the time for the [tex]i[/tex]th task.
We know these times have a certain structure:
Any 3 adjacent tasks will take half as long as the next two tasks.In the form of an equations we have
[tex]t_1+t_2+t_3=\frac{1}{2}t_4+\frac{1}{2}t_5 \\\\t_2+t_3+t_4=\frac{1}{2}t_5+\frac{1}{2}t_6[/tex]
The second task takes 1 second [tex]t_2=1[/tex]The fourth task takes 10 seconds [tex]t_4=10[/tex]So, we have the following system of equations:
[tex]t_1+t_2+t_3-\frac{1}{2}t_4-\frac{1}{2}t_5=0 \\\\t_2+t_3+t_4-\frac{1}{2}t_5-\frac{1}{2}t_6=0\\\\t_2=1\\\\t_4=10[/tex]
a) An augmented matrix for a system of equations is a matrix of numbers in which each row represents the constants from one equation (both the coefficients and the constant on the other side of the equal sign) and each column represents all the coefficients for a single variable.
Here is the augmented matrix for this system.
[tex]\left[ \begin{array}{cccccc|c} 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 & 0 \\\\ 0 & 1 & 1 & 1 & - \frac{1}{2} & - \frac{1}{2} & 0 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]
b) To reduce this augmented matrix to reduced echelon form, you must use these row operations.
Subtract row 2 from row 1 [tex]\left(R_1=R_1-R_2\right)[/tex].Subtract row 2 from row 3 [tex]\left(R_3=R_3-R_2\right)[/tex].Add row 3 to row 2 [tex]\left(R_2=R_2+R_3\right)[/tex].Multiply row 3 by −1 [tex]\left({R}_{{3}}=-{1}\cdot{R}_{{3}}\right)[/tex].Add row 4 multiplied by [tex]\frac{3}{2}[/tex] to row 1 [tex]\left(R_1=R_1+\left(\frac{3}{2}\right)R_4\right)[/tex].Subtract row 4 from row 3 [tex]\left(R_3=R_3-R_4\right)[/tex].Here is the reduced echelon form for the augmented matrix.
[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \end{array} \right][/tex]
c) The additional rows are
[tex]\begin{array}{ccccccc} 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right[/tex]
and the augmented matrix is
[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & \frac{1}{2} & 15 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & - \frac{1}{2} & - \frac{1}{2} & -11 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \\\\ 1 & 1 & 1 & 0 & 0 & 0 & 50 \end{array} \right][/tex]
d) To solve the system you must use these row operations.
Subtract row 1 from row 6 [tex]\left(R_6=R_6-R_1\right)[/tex].Subtract row 2 from row 6 [tex]\left(R_6=R_6-R_2\right)[/tex].Subtract row 3 from row 6 [tex]\left(R_6=R_6-R_3\right)[/tex].Swap rows 5 and 6.Add row 5 to row 3 [tex]\left(R_3=R_3+R_5\right)[/tex].Multiply row 5 by 2 [tex]\left(R_5=\left(2\right)R_5\right)[/tex].Subtract row 6 multiplied by 1/2 from row 1 [tex]\left(R_1=R_1-\left(\frac{1}{2}\right)R_6\right)[/tex].Add row 6 multiplied by 1/2 to row 3 [tex]\left(R_3=R_3+\left(\frac{1}{2}\right)R_6\right)[/tex].[tex]\left[ \begin{array}{ccccccc} 1 & 0 & 0 & 0 & 0 & 0 & 5 \\\\ 0 & 1 & 0 & 0 & 0 & 0 & 1 \\\\ 0 & 0 & 1 & 0 & 0 & 0 & 44 \\\\ 0 & 0 & 0 & 1 & 0 & 0 & 10 \\\\ 0 & 0 & 0 & 0 & 1 & 0 & 90 \\\\ 0 & 0 & 0 & 0 & 0 & 1 & 20 \end{array} \right][/tex]
The solutions are: [tex](t_1,...,t_6)=(5,1,44,10,90,20)[/tex].
The intelligence quotients (IQs) of 16 students from one area of a city showed a mean of 107 and a standard deviation of 10, while the IQs of 14 students from another area of the city showed a mean of 112 and a standard deviation of 8. Is there a significant difference between the IQs of the two groups at significance level of 0.01. What is the alternative hypothesis?
Answer:
No, there is no significant difference between the IQs of the two groups.Alternative Hypothesis is that the two groups have different IQs os students.Step-by-step explanation:
We are provided that IQs of 16 students from one area of a city had a mean of 107 and a standard deviation of 10 while the IQs of 14 students from another area of the city had a mean of 112 and a standard deviation of 8.
And we have to check that is there a significant difference between the IQs of the two groups.
Firstly let, Null Hypothesis, [tex]H_0[/tex] : The two groups have same IQs { [tex]\mu_1 = \mu_2[/tex] }
Alternate Hypothesis, [tex]H_1[/tex] : The two groups have different IQs{ [tex]\mu_1 \neq \mu_2[/tex]}
Since we don't know about population standard deviations;
The test statistics we will use here will be ;
[tex]\frac{(X_1bar - X_2bar)- (\mu_1 - \mu_2) }{s_p\sqrt{\frac{1}{n_1}+\frac{1}{n_2} } }[/tex] follows t distribution with [tex](n_1 + n_2 -2)[/tex] degree
of freedom { [tex]t_n__1 +n_2 - 2[/tex] }
Here, [tex]X_1bar[/tex] = 107 [tex]X_2bar[/tex] = 112 [tex]s_1[/tex] = 10 [tex]s_2[/tex] = 8
[tex]n_1[/tex] = 16 [tex]n_2[/tex] = 14
[tex]s_p[/tex] = [tex]\sqrt{\frac{(n_1 - 1)*s_1^{2} + (n_2 -1)*s_2^{2} }{(n_1 + n_2 -2)} }[/tex] = 9.1261
Test statistics = [tex]\frac{(107-112) - 0}{9.1261*\sqrt{\frac{1}{16} +\frac{1}{14} } }[/tex] follows [tex]t_2_8[/tex]
= -1.50
Now at 1% level of significance t table is giving the critical value of -2.467 and our test statistics is higher than this means it does not fall in the rejection region so we will accept our null hypothesis and conclude that there is no significant difference between the IQs of the two groups.
This question can be addressed by conducting a two-sample t-test to determine if there is a significant difference between the mean intelligence quotients of students from two areas of a city. The steps include calculating pooled standard deviation, followed by standard error, calculating the t-score, and comparing it to a critical value. You can either reject or fail to reject the null hypothesis based on these results.
Explanation:The question is asking if there is a significant difference in the mean intelligence quotients (IQs) of students between two areas of a city. Conducting a two-sample t-test can address this. First, let's define the null and alternative hypotheses.
Null Hypothesis (H₀): There is no significant difference between the two sets of IQ scores (mean1 = mean2).Alternative Hypothesis (Hᵃ): There is a significant difference between the two sets of IQ scores (mean1 ≠ mean2).To perform this test, follow these steps:
Calculate the pooled standard deviation for the two samples.Use it to determine the standard error of the difference between the two means.Use these to calculate the t-score.Compare the t-score with the critical value for the t-distribution with a significance level of 0.01. If the t-score exceeds the critical value, you reject the null hypothesis, i.e., there's a significant difference in IQs between the areas of the city. While if it is less, the null hypothesis is not rejected, thus, no significant difference.Remember, depending on the specific results, we may or may not find enough evidence to support the alternative hypothesis.
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What does the term "expand" mean in mathematics?
I am NOT searching for "expanded form" or "distribute".
I think expanding means to remove the parentheses/brackets from a problem.
For example: Say we have the expression: 3 (4 + 5). I think expanding means to multiply 3, by every number in the parentheses. So that means:
(3 * 4) + (3 * 5) = 27.
Another way to think about it is to (if you're on paper) draw a line from 3, to all the numbers inside the parentheses. The line that connects from 3 to 4, is signaling for you to multiply 3 * 4 = 12. And the line from 3 to 5 = 3 * 5 = 15. And add them.
Final answer:
In mathematics, 'expand' refers to writing an expression in an extended form using distribution. This can result in a polynomial or an infinite series, as seen in binomial expansion or exponential arithmetic.
Explanation:
In mathematics, to expand means to increase the length of an expression by distributing multiplication over addition or subtraction. For example, expanding (a + b)(c + d) results in ac + ad + bc + bd. This does not change the value of the expression, but rather writes it in an alternative form that might be more useful for further operations, such as simplification or evaluation. Binomial expansion, specifically, refers to expressing a binomial raised to a power as a series of terms, using the binomial theorem, which can sometimes result in an infinite series or a polynomial of finite length. This expansion is applicable in situations like expanding (x + y)^n or when dealing with power series expansions of standard mathematical functions including exponential arithmetic where numbers are expressed as a product of a digit term and an exponential term such as in the notation 4.57 x 10^3.
Consider the baggage check-in process of a small airline. Check-in data indicate that from 9 a.m. to 10 a.m., 220 passengers checked in. Moreover, based on counting the number of passengers waiting in line, airport management found that the average number of passengers waiting for check-in was 27.?How long did the average passenger have to wait in line?
Answer:
The passengers have an average of 8.15 minutes to wait in line
Step-by-step explanation:
Using Little's law
Average Inventory = Average Flow Time * Average Flow Rate
Average Inventory = 220 passengers
Average Flow Rate = 27
Average Flow time =?
So,
220 = Average Flow Time * 27
Average Flow Time = 220/27
Average Flow Time = 8.14814814814
Average Flow Time = 8.15 --------- Approximated
So the average wait time for a passenger is 8.15 minutes
The average passenger had to wait in line for approximately 0.123 minutes.
Explanation:To find the average waiting time, we need to divide the total waiting time by the number of passengers. The total waiting time can be calculated by multiplying the average number of passengers waiting (27) by the time period (1 hour). The average waiting time is then found by dividing this total waiting time by the number of passengers checked in (220).
Average waiting time = (Average number of passengers waiting × Time period) / Number of passengers checked in
First, find the total waiting time: Total waiting time = 27 × 1 = 27 minutesNext, find the average waiting time: Average waiting time = 27 / 220 = 0.123 minutes (approximately)Therefore, the average passenger had to wait in line for approximately 0.123 minutes.
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Abby is buying a widescreen TV that she will hang on the wall between two windows. The windows are 36 inches apart, and wide screen TVs are approximately twice as wide as they are tall. Of the following, which is the longest that the diagonal of a widescreen TV can measure and still fit between the windows
Answer:
D < 40.2 inches
Step-by-step explanation:
The maximum width of the TV must be 36 inches. Since TVs are approximately twice as wide as they are tall, the maximum height is 18 inches.
The diagonal of a TV can be determined as a function of its width (w) and height (h) as follows:
[tex]d^2=h^2+w^2\\d=\sqrt{18^2+36^2}\\d= 40.2\ in[/tex]
Therefore, the diagonal must be at most 40.2 inches.
Since the answer choices were not provided with the question, you should choose the biggest value that is under 40.2 inches.
The maximum diagonal size of the widescreen TV that can fit between two windows 36 inches apart is slightly more than 40 inches, given that the TV has an aspect ratio where the width is about twice the height.
Let's denote the TV's width as w and the height as h. Given that widescreen TVs are about twice as wide as they are tall, we can express the width as w = 2h.
The diagonal d of the TV can be found using Pythagoras' theorem where d² = w² + h².
Substituting 2h for w, we get d² = (2h)² + h² which simplifies to d² = 4h² + h² and further to d² = 5h².
Thus, d = h√5.
If the space between windows is 36 inches, this would be the maximum width of the TV. Therefore, 36 = 2h which means that h = 18 inches. Using this height in the diagonal equation, we get d = 18√5 which is approximately 40.2 inches. This means the longest diagonal of the widescreen TV that can fit between the windows is slightly more than 40 inches.
Recursive definitions for subsets of binary strings.Give a recursive definition for the specified subset of the binary strings. A string r should be in the recursively defined set if and only if r has the property described. The set S is the set of all binary strings that are palindromes. A string is a palindrome if it is equal to its reverse. For example, 0110 and 11011 are both palindromes.
Answer:
Step-by-step explanation:
A binary string with 2n+1 number of zeros, then you can get a binary string with 2n(+1)+1 = 2n+3 number of zeros either by adding 2 zeros or 2 1's at any of the available 2n+2 positions. Way of making each of these two choices are (2n+2)22. So, basically if b2n+12n+1 is the number of binary string with 2n+1 zeros then your
b2n+32n+3 = 2 (2n+2)22 b2n+12n+1
your second case is basically the fact that if you have string of length n ending with zero than you can the string of length n+1 ending with zero by:
1. Either placing a 1 in available n places (because you can't place it at the end)
2. or by placing a zero in available n+1 places.
0 ϵ P
x ϵ P → 1x ϵ P , x1 ϵ P
x' ϵ P,x'' ϵ P → xx'x''ϵ P
A recursive definition for the set of binary string palindromes starts with the base cases '0' and '1'. Other palindromes can be obtained by nesting a palindrome between '0' and '0' or '1' and '1'.
A recursive definition for the set S, consisting of all binary strings that are palindromes, would be defined by two rules:
For the base cases, both '0' and '1' are in S. This covers the palindromes of length 1.
The inductive step would be: If 'P' is a string in S, then both '0P0' and '1P1' are in S. This allows us to generate palindromes of increasing lengths all the way to infinity.
By this definition, a string is a palindrome if it is the same when read from left to right and right to left. It starts with the simplest cases (single digit palindromes) and then defines how to build larger examples based on smaller ones.
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Based on a Pitney Bowes survey, assume that 42% of consumers are comfortable having drones deliver their purchases. Suppose we want to find the probability that when five consumers are randomly selected, exactly two of them are comfortable with the drones. What is wrong with using the multiplication rule to find the probability of getting two consumers comfortable with drones followed by three consumers not comfortable, as in this calculation: 10.42210.42210.58210.58210.582 = 0.0344?
Answer:
For this case is wrong use the multiplication for P(X=2):
0.42*0.42*0.58*0.58*0.58 = 0.0344
Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:
[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]
And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.
Step-by-step explanation:
Previous concepts
The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".
Solution to the problem
Let X the random variable of interest, on this case we now that:
[tex]X \sim Binom(n=5, p=0.42)[/tex]
The probability mass function for the Binomial distribution is given as:
[tex]P(X)=(nCx)(p)^x (1-p)^{n-x}[/tex]
Where (nCx) means combinatory and it's given by this formula:
[tex]nCx=\frac{n!}{(n-x)! x!}[/tex]
For this case is wrong use the multiplication for P(X=2):
0.42*0.42*0.58*0.58*0.58 = 0.0344
Because we don't take in count the possible nCx ways in order to have the two consumers comfortable, and we are assuming that the first two people are comfortable and the rest is not, and that's not the only possibility. The correct probability for X=2 people comfortable is given by:
[tex]P(X=2)=(5C2)(0.42)^2 (1-0.42)^{5-2}=0.344[/tex]
And as we can see the real answer is 10 times the assumed answer, for this reason is wrong the claim.
special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin. A. How many elements are there in the sample space? B. Let A be the event that a red card is picked first, followed by landing a tail on the coin toss. P(A) = Present your answer as a decimal number to 1 decimal place. C. Let B be the event that a green or blue is picked, followed by landing a tail on the coin toss. Are the events A and B mutually exclusive? D. Let C be the event that a green or red is picked, followed by landing a tail on the coin toss. Are the events A and C mutually exclusive?
Answer:
Step-by-step explanation:
Given that special deck of cards has 20 cards. Nine are green, seven are blue, and four are red. When a card is picked, the color of it is recorded. An experiment consists of first picking a card and then tossing a coin
A) Sample space will have Green, Head, or Green, Tail .... Red, head, red, tail
No of elements in sample space = no of colours x no of outcomes in coin toss
= 4x2 = 8
B) A= getting (RT)
P(A) = Prob of getting red card and tail on coin
= P (R) *P(T)
=[tex]\frac{4}{20} *\frac{1}{2} \\=\frac{1}{10}[/tex]
C) B be the event that a green or blue is picked, followed by landing a tail on the coin toss
B = getting green card and tail
Getting green card tail is mutally exclusive with red card and tail as there is no common element between green and blue.
D) C= red or green card is picked followed by tail.
Here A and C have a common element as getting red and tail. So not mutually exclusive
Final answer:
The sample space of the card and coin toss experiment consists of 40 elements. The probability of picking a red card followed by a tail coin toss is 0.1. Events A and B, as well as events A and C, are not mutually exclusive.
Explanation:
Probability in a Card and Coin Toss Experiment
When dealing with a special deck of cards and a subsequent coin toss, multiple steps are involved in determining outcomes and probabilities. As for the given student's question:
A). The sample space contains multiple elements based on the card colors and the side of the coin: each of the 20 cards can result in either heads or tails, creating a total of 40 possible outcomes.
B). For event A, where a red card is picked followed by a tails on the coin toss, the probability (P(A)) is calculated by dividing the number of successful outcomes by the number of total possibilities, resulting in P(A) = 4 (red cards) * 1 (tails outcome) / 40 (total outcomes) = 0.1.
C). Event A and event B are not mutually exclusive because event B involves picking either a green or blue card and also getting tails, which does not overlap with the specifics of event A.
D). Similarly, events A and C are not mutually exclusive. Although both involve picking a red card and landing a tail, event C also includes picking a green card, which does not interfere with the occurrence of event A.
1) Let f(x)=ax2+bx+c for some value of a, b and c. f intersects the x-axis when x=−2 or x=3, and f( 1 3 )=−25. Find the values of a, b and c and sketch the graph of f(x).
2) A right prism has a base that is an equilateral triangle. The height of the prism is equal to the height of the base. If the volume of the prism is 81, what are the lengths of the sides of the base?
thank u sm
Answer:
1) a = -⅙, b = ⅙, c = 1
2) 6 units
Step-by-step explanation:
1) f(x) = ax² + bx + c
Given the roots, we can write this as:
f(x) = a (x + 2) (x − 3)
We know that f(13) = -25, so we can plug this in to find a:
-25 = a (13 + 2) (13 − 3)
-25 = 150a
a = -⅙
Therefore, the factored form is:
f(x) = -⅙ (x + 2) (x − 3)
Distributing:
f(x) = -⅙ (x² − x − 6)
f(x) = -⅙ x² + ⅙ x + 1
Graph: desmos.com/calculator/6m6tjoodvb
2) Volume of a right prism is area of the base times the height.
V = Ah
The base is an equilateral triangle. Area of a triangle is one half the base times height.
V = ½ ab h
The height of the triangle is the same as the height of the prism.
V = ½ bh²
In an equilateral triangle, the height is equal to half the base times the square root of 3.
V = ½ b (½√3 b)²
V = ⅜ b³
Given that V = 81, solve for b.
81 = ⅜ b³
216 = b³
b = 6
. During a year, the probability a structure will be damaged by an earthquake (A) is 0.02, that it will be damaged by a hurricane (B) is 0.03, and that it will be damaged by both is 0.007. What is the probability that it will not be damaged by a hurricane or an earthquake during that year?
Answer:
0.9506
Step-by-step explanation:
Pr(A) = 0.02
Pr(B) = 0.03
Pr(both) = 0.007
So,
Pr(Not A) = 1 - Pr(A)
= 1 - 0.02
= 0.98
Pr(Not B) = 1 - Pr(B)
= 1 - 0.03
= 0.97
Pr(Not by both) = 1 - Pr(both)
= 1 - 0.007
= 0.993
Thus,
Pr(Not B) or Pr(Not A) = 0.97 × 0.98
= 0.9056
∴ the probability that the house would not be damaged by a hurricane or an earthquake during the year is 0.9506.
How many pounds of oranges do the data in the plot line represent?
Answer:
OPTION C: [tex]$ \textbf{37} \frac{\textbf{28}}{\textbf{8}} $[/tex] pounds.
Step-by-step explanation:
From the figure we can see that there are three dots against [tex]$ 3 \frac{7}{8} $[/tex].
That means it becomes [tex]$ 3 \times 3\frac{7}{8} $[/tex].
Note that if there is a mixed fraction of the form [tex]$ a \frac{b}{c} $[/tex] = [tex]$ a + \frac{b}{c} $[/tex].
Therefore, [tex]$ 3 \times 3\frac{7}{8} = 3 \times \bigg(3 + \frac{7}{8} \bigg ) $[/tex] ... (1)
Similarly, against 4 there are 2 dots.
So, it should be [tex]$ 4 \times 2 $[/tex] pounds. ...(2)
3 dots against [tex]$ 4 \frac{1}{8} $[/tex].
So, it becomes [tex]$ 3 \times \bigg(4 + \frac{1}{8} \bigg) $[/tex] ...(3)
Similarly, 2 dots against [tex]$ 4 + \frac{2}{8} $[/tex].
This will become [tex]$ 2 \times \bigg( 2 + \frac{2}{8} \bigg) $[/tex] ...(4)
Now, to calculate the total pound, we simply add (1), (2), (3) & (4).
⇒ [tex]$ 3 \times \bigg(3 + \frac{7}{8} \bigg ) $[/tex] [tex]$ + $[/tex] [tex]$ 4 \times 2 $[/tex] + [tex]$ 3 \times \bigg(4 + \frac{1}{8} \bigg) $[/tex] [tex]$ + $[/tex] [tex]$ 2 \times \bigg( 2 + \frac{2}{8} \bigg) $[/tex]
⇒ [tex]$ 9 + \frac{21}{8} + 8 + 12 + \frac{3}{8} + 8 + \frac{4}{8} $[/tex]
⇒ [tex]$ \bigg ( 9 + 8 + 12 + 8 \bigg) + \bigg( \frac{21 + 3 + 4}{8} \bigg ) $[/tex]
⇒ [tex]$ \textbf{37} \textbf {+} \frac{\textbf{28}}{\textbf{8}} $[/tex] [tex]$ \textbf{=} \hspace{1mm} \textbf{37} \frac{\textbf{28}}{\textbf{8}} $[/tex] which is the required answer.
Suppose that diameters of a new species of apple have a bell-shaped distribution with a mean of 7.25 cm and a standard deviation of 0.42 cm. Using the empirical rule, what percentage of the apples have diameters that are between 6.41cm and 8.09 cm?
Answer:
95%
Step-by-step explanation:
Upper limit = 8.09 cm
Lower limit = 6.41 cm
Distribution mean = 7.25 cm
Standard deviation = 0.42 cm
The number of standard deviations from the mean of the upper and lower limits are, respectively:
[tex]N_U=\frac{U-M}{SD} =\frac{8.09-7.25}{0.42}=2 \\N_L=\frac{M-L}{SD} =\frac{7.25-6.41}{0.42}=2[/tex]
Both limits are two standard deviations away from the mean.
According to the empirical rule, in normal distributions, 95% of the data falls within two standard deviations of the mean. Therefore, 95% of the apples have diameters that are between 6.41cm and 8.09 cm.
According to the National Household Survey on Drug Use and Health, when asked in 2012, 41% of those aged 18 to 24 years used cigarettes in the past year, 9% used smokeless tobacco, 36.3% used illicit drugs, and 10.4% used pain relievers or sedatives. Explain why it is not correct to display these data in a pie chart.
a. The types of illicit drugs are not given.
b. There could be roundoff error.
c. The three groups do not add up to 100%.
d. There have to be more than three categorical variables.
e. There could be overlap between the groups.
Final answer:
The correct answer is c. The three groups do not add up to 100%. A pie chart is used to display the parts of a whole, where each category represents a proportion of the total. In this case, the categories of cigarette use, smokeless tobacco use, illicit drug use, and pain reliever/sedative use do not add up to 100% when combined. As a result, a pie chart would not accurately represent the data.
Explanation:
The correct answer is c. The three groups do not add up to 100%.
A pie chart is used to display the parts of a whole, where each category represents a proportion of the total. In this case, the categories of cigarette use, smokeless tobacco use, illicit drug use, and pain reliever/sedative use do not add up to 100% when combined. As a result, a pie chart would not accurately represent the data.
Calculate the sample standard deviation and sample variance for the following frequency distribution of heart rates for a sample of American adults. If necessary. round to one more decimal place than the largest number of decimal places given in the data. Heart Rates in Beats per Minute Class Frequency 61-6613 67-72 10 73-78 3 79-8411 85-90 3
Answer:
[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]
[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]
[tex] s = \sqrt{70.131}= 8.374[/tex]
Step-by-step explanation:
For this case we can calculate the expected value with the following table"
Class Midpoint(xi) Freq. (fi) xi fi xi^2 * fi
61-66 63.5 13 825.5 52419.5
67-72 69.5 10 695 48302.5
73-78 75.5 3 226.5 17100.75
79-84 81.5 11 896.5 73064.75
85-90 87.5 3 262.5 22968.75
________________________________________________
Total 40 2906 213856
For this case the midpoint is calculated as the average between the minimum and maximum point for each class.
The expected value can be calculated with the following formula:
[tex] \bar X = \frac{\sum_{i=1}^5 x_i f_i}{n} = \frac{2906}{40}= 72.65[/tex]
For this case n =40 represent the total number of obervations given,
And for the sample variance we can use the following formula:
[tex] s^2 = \frac{\sum x^2_i f_i -\frac{\sum x_i f_i}{n}}{n-1}[/tex]
And if we replace we got:
[tex] s^2 = \frac{213856 -\frac{2906}{40}}{40-1}=70.131[/tex]
And for the deviation we take the square root:
[tex] s = \sqrt{70.131}= 8.374[/tex]
To calculate the sample standard deviation and sample variance, first calculate the sample mean, then calculate the sample variance, and finally find the square root of the sample variance to get the sample standard deviation.
Explanation:To calculate the sample standard deviation and sample variance for the given frequency distribution of heart rates, we need to follow these steps:
Create a chart to organize the data, frequencies, relative frequencies, and cumulative relative frequencies. Calculate the sample mean (average) by multiplying each heart rate value by its frequency, summing those products, and dividing by the total number of observations. Calculate the sample variance by finding the squared difference between each heart rate value and the mean, multiplying each squared difference by its frequency, summing those products, and dividing by the total number of observations minus 1. Calculate the sample standard deviation by taking the square root of the sample variance.
Using the provided data, the sample standard deviation and sample variance can be calculated as follows:
Sample mean = (65 * 13 + 69.5 * 10 + 75.5 * 3 + 81.5 * 11 + 87.5 * 3) / (13 + 10 + 3 + 11 + 3) ≈ 72.74
Sample variance = [(65 - 72.74)² * 13 + (69.5 - 72.74)² * 10 + (75.5 - 72.74)² * 3 + (81.5 - 72.74)² * 11 + (87.5 - 72.74)² * 3] / (13 + 10 + 3 + 11 + 3 - 1) ≈ 54.21
Sample standard deviation = √(54.21) ≈ 7.36
the average cost of living in san francisco?
Step-by-step explanation:
The median rent for a one-bedroom apartment stands at $3,460 a month.
Also The estimated cost of annual necessities for a single person is $43,581 — or $3,632 a month, making it the most expensive city for single people to settle down in.
And For a family of four, expect to pay about $91,785 a year for necessities — that's $7,649 per month.
For a family of four, expect to pay about $91,785 a year for necessities — that's $7,649 per month.
Although exact data isn't provided, information on related costs such as average salary and gasoline prices suggest that the average cost of living in San Francisco is high.
Explanation:The average cost of living in San Francisco is significantly higher than the national average. According to Numbeo, San Francisco's overall cost of living index is 176.89, which is 76.89% higher than the U.S. average of 100. This means that you can expect to pay about 77% more for goods and services in San Francisco than you would in the average American city.
However, we can infer that the cost of living is high, considering the mean starting salary for San Jose State University graduates, nearby to San Francisco, is at least $100,000 per year. This suggests that a significant income is required to support oneself in the Bay Area.
Other factors indirectly hint at the costs associated with San Francisco living. For instance, the average cost of unleaded gasoline in the Bay Area was once $4.59, which is notably high. These pieces of information, though incomplete, indicate a high cost of living.
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How does hypothesis testing differ from constructing confidence intervals, in general? Read carefully.
Answer:
Step-by-step explanation:
Hypothesis testing invariably used to test a claim about a population parameter is widely used to check whether they hypothetical claim made is right.
For example, mean scores of a particular college is more than 75% is tested with hypothesis as setting null as equal to 75% and alternate >75%
Processes are done stepwise from the sample collected and conclusion made
Confidence interval on the other hand is the range of values within which the parameter is expected to lie at a certain confidence level
Estimated population parameter is provided for error known as margin of error depending upon the confidence level, and an interval is prepared which guarantees to the extent of confidence that parameters will fall within.
Hypothesis testing can be concluded with the use of confidence intervals also.
Calculate descriptive statistics for the variable (Coin) where each of the thirty-five students in the sample flipped a coin 10 times. Round your answers to three decimal places and write the mean and the standard deviation.
Answer:
Mean = 5; Standard Deviation: 1.5811
Step-by-step explanation:
Given Data:
number of times coin flipped = n = 10;
probability of each side of coin = p = 0.5;
Here mean is the product of number of times coin flipped and probability of each
m = n*p =10*0.5 = 5
Standard deviation is obtained by taking square root of product of n,p,q
St. Dev= [tex]\sqrt{npq}[/tex] = [tex]\sqrt{10*0.5*0.5}[/tex] = 1.5811
We have:
Mean = 5 ; Standard Deviation = 1.5811
Write down the general zeroth order linear ordinary differential equation. Write down the general solution.
The zeroth derivative of a function [tex]y(x)[/tex] is simply the function itself, so the zeroth order linear ODE takes the general form
[tex]y(x)=f(x)[/tex]
whose solution is [tex]f(x)[/tex].