Although the rules of probability are just basic facts about percents or proportions, we need to be able to use the language of events and their probabilities. Choose an American adult aged 20 20 years and over at random. Define two events: A = A= the person chosen is obese B = B= the person chosen is overweight, but not obese

Answer:

37 and 26

Step-by-step explanation:

Let the bigger number be x and the smaller number be y.

Then the two numbers total 63 will give us:

y+x=63------>eqn1

The two numbers have a difference of 11.

y-x=11------->eqn2

We add equations 1 and 2 to obtain:

2y=74

Divide both sides to get:

y=37

Put y=37 into eqn2 to get:

37-x=11

x=37-11

x=26

This larger number is 37 and the smaller number is 26

Answer:

Step-by-step explanation:

given is a vector as (5,9,3)

a = (5,9,3)

To find out direction cosines

First let us calculate modulus of vector a

[tex]||a|| =\sqrt{5^2+9^2+3^2} \\=\sqrt{25+81+9} \\=\sqrt{115}[/tex]

Direction ratios are (5,9,3)

Magnitude of vector a = [tex]\sqrt{115}[/tex]

So direction cosines would be

[tex](\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

Angles would be

[tex](\alpha, \beta, \gamma) = arccos ((\frac{5}{\sqrt{115} } ,\frac{9}{\sqrt{115} },\frac{3}{\sqrt{115} })[/tex]

=cos inverse  (0.4662, 0.8393, 0.2798)

= (62.21, 32.93,32,94)

The two vectors which are orthogonal to the given vector u are (0, 4, 1) and  (8, 0, -5).

Use the concept of orthogonal vectors defined as:

The term "orthogonal" in mathematics denotes a direction at a 90° angle. If two vectors u, and v form a right angle when they are perpendicular, or if the dot product they produce is zero, then they are said to be orthogonal.

The given vector is,

u = (5, −2, 8)

Let the vector orthogonal to this is,

v = (x, y, z)

Since vectors u and v are orthogonal,

Therefore,

u.v = 0

(5, −2, 8)(x, y, z) = 0

5x -2y + 8x = 0  .....(i)

To find the vectors choose x, y, and x such that equation one is satisfied.

So take x = 0, y = 4 and z = 1

Then (x, y, z) = (0, 4, 1) satisfy the equation (i)

Similarly, choose x = 8, y = 0, z = -5

Then (x, y, z) = (8, 0, -5) satisfy the equation (i)

Hence,

Two orthogonal vectors are (0, 4, 1) and  (8, 0, -5)

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Two vectors orthogonal to u=(5, -2, 8) could be v=(2, 5, 0) and u'=(-2, -5, 0). Both vectors satisfy the definition of orthogonality, having a dot product with u equal to zero.

To find two vectors that are orthogonal to the given vector u = (5, -2, 8), we need to find two vectors that have a dot product with vector u equal to zero. This is because orthogonality (perpendicularity in 3D space) is defined by a zero dot product.

For instance, let's calculate the dot product of u with v = (2, 5, 0). It'll be (5*2) + ((-2)*5) + (8*0) = 0. Therefore, vector v = (2, 5, 0) is orthogonal to u.

For a vector in the opposite direction, we simply need to multiply every component of vector v by -1. Consequently, a vector u′ = (-2, -5, 0) is in the opposite direction of v but still orthogonal to u. So, our answer is vectors v = (2, 5, 0) and u' = (-2, -5, 0).

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Answer:

For the exhange of ions and gases such as potassium, calcium and sodium, large fish larva have gills even in early stages. One other main reason is that large fish arvas mostly have circulating red cells after hatching phase while small fish larvae doesn't. So, for these reasons small fish larvae have no gills until later in life.

Answer:

The answer to your question is

Step-by-step explanation:

Inequality 1

                             4(3x - 4) < 32

                             12x - 16 < 32

                             12x < 32 + 16

                              12x < 48

                                 x < 48/12

                                 x < 4

Inequality 2

                           2x + 1 ≤ 8x + 25

                           2x - 8x ≤ 25 - 1

                               - 6x ≤ 24

                                   x ≥ 24/-6

                                   x ≥ - 4

- See the graph below

- Interval [-4, 4)

To solve the inequality system, we divide both sides of each inequality by the respective coefficient to isolate the variable. The solutions are x < 4 and -4 ≤ x. The graph of the solution is a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

To solve the inequality 4(3x - 4) < 32, we divide both sides of the inequality by 4 to isolate the variable. This gives us 3x - 4 < 8. Adding 4 to both sides of the inequality gives us 3x < 12.

Finally, dividing both sides of the inequality by 3 gives us x < 4.

For the inequality 2x + 1 ≤ 8x + 25, we subtract 2x from both sides of the inequality to isolate the variable. This gives us 1 ≤ 6x + 25. Subtracting 25 from both sides of the inequality gives us -24 ≤ 6x.

Finally, dividing both sides of the inequality by 6 gives us -4 ≤ x.

The solutions to the inequality system are x < 4 and -4 ≤ x. The graph of the solution would be a number line with an open circle at 4 and a shaded region to the left, and a closed circle at -4 and a shaded region to the right.

The interval notation for the solution is (-∞, 4) and [-4, ∞).

Answer:

The remainder is 9

Step-by-step explanation:

3^128 is divided by 17

find the value of 3^128 first.

3^128 = 1.17901845777E61

Then you divide by 17

1.17901845777E61 ÷ 17

= 6.93540269279E59

Approximately 6. 9340

6 remainder 9

Answer:

6 remander 1

Step-by-step explanation:

first you solve 3^128 which equals E61 then divide it by 17 which equals E59

Answer:

The mean number of adults who would have bank savings accounts is 32.

Step-by-step explanation:

For each adult surveyed, there are only two possible outcomes. Either they have bank savings accounts, or they do not. So we use the binomial probability distribution to solve this problem.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

The expected value of the binomial distribution is:

[tex]E(X) = np[/tex]

In this problem, we have that:

[tex]p = 0.64[/tex]

If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

This is E(X) when [tex]n = 50[/tex].

So

[tex]E(X) = np = 50*0.64 = 32[/tex]

The mean number of adults who would have bank savings accounts is 32.

You can convert the given normal distribution to standard normal distribution and then use the z tables to find the needed probabilities.

Rounding to one places of decimal, we get the answers as:

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

If we have

[tex]X \sim N(\mu, \sigma)[/tex]

(X is following normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex])

then it can be converted to standard normal distribution as

[tex]Z = \dfrac{X - \mu}{\sigma}, \\\\Z \sim N(0,1)[/tex]

Using the z scores will help to find the probabilities from the z tables(available online).

Let for the given test, the test scores be tracked by a random variable X, then by the given data, we have:

[tex]X \sim N(530, 119)[/tex]

The needed probabilities are

Converting the distribution to standard normal variate, the probabilities become

[tex]Z = \dfrac{X - 530}{119}\\\\Z \sim N(0, 1)[/tex]
The probabilities convert to

a) [tex]P( 411 < X < 649 ) = P(X < 649) - p(X < 411)\\[/tex]

[tex]P(\dfrac{411 - 530}{119} < Z < \dfrac{ 649 - 530}{119}) = P(-1 < Z < 1) = P(Z < 1) - P(Z < -1)[/tex]

(Know the fact that in continuous distribution, probability of a single point is 0, so we can write [tex]P(Z < a) = P(Z \leq a)[/tex] )

Also, know that if we look for Z = a in z tables, the p value we get is [tex]P(Z \leq a) = p \: value[/tex]

The p value at Z = 1 is 0.8413 and at Z = -1 is 0.1587,

Thus, [tex]P(411 < X < 649) = P(Z < 1) - P(Z < -1) = 0.8413 - 0.1587 = 0.6826[/tex]

b) [tex]P(X < 411) + P(X > 649) = 1 - P(411 \leq X \leq 649) = 1 - P(411 < X < 649)\\\\P(X < 411) + P(X > 649) = 1 - 0.6826 = 0.3174[/tex]

c) [tex]P(X > 768) = 1 - P(X \leq 768)[/tex]

[tex]P(X > 768) = 1 - P(Z < \dfrac{768 - 530}{119}) = 1 - P(Z < 2) = 1 - 0.9772 = 0.0228[/tex]

Rounding to one places of decimal, we get the answers as:

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The volume of dump truck is 237.7 feet cubed.

Step-by-step explanation:

Given dimensions are;

Length of dump truck = 12.05 feet

Width of dump truck = 3.86 feet

Height of dump truck = 5.11 feet

Volume = Length * Width * Height

Volume = 12.05 * 3.86 * 5.11

Volume = 237.681 ft³

Rounding off to nearest tenth

Volume = 237.7 ft³

The volume of dump truck is 237.7 feet cubed.

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Answer:

a.) A= [RRR, LLL, SSS]

b.) B=[RLS, RSL, SRL, SLR, LRS, LSR]

c.) C=[RRL, RRS, RSR, RLR, SRR, LRR]

e.) D'= C' = [RRR, LLL, SSS, RLS, RSL, SRL, SLR, LRS, LSR, LLR, LLS, LSL, LRL, SLL, RLL, SSL, SSR, SLS, SRS, RSS, LSS]

Step-by-step explanation:

If three consecutive vehicles are considered with their direction as R, L and S, then

a.) when the three vehicles go in same direction, A= [RRR, LLL, SSS]

b.) When the three vehicles take the different direction, B=[RLS, RSL, SRL, SLR, LRS, LSR]

c.) when exactly two vehicles turn right, C=[RRL, RRS, RSR, RLR, SRR, LRR]

d.) repeated question of C above.

e.) D'= C' = [RRR, LLL, SSS, RLS, RSL, SRL, SLR, LRS, LSR, LLR, LLS, LSL, LRL, SLL, RLL, SSL, SSR, SLS, SRS, RSS, LSS]

Answer:

Let G be the number of eggs in the meal

V be the number of servings of mixed vegetables in the meal

R be the number of servings of brown rice in the meal

Objective function = Minimize 3.75G + 3.50V + 2R

Constraints:

2G + 14V + 40R ≥ 35(Carbohydrates)

18G + 15V + 6R ≥ 30(Protein)

12G + 8V + R ≤ 45(Fat)

G, M, B ≥ 0

Answer:

[tex]P(X<480)=P(\frac{X-\mu}{\sigma}<\frac{480-\mu}{\sigma})=P(Z<\frac{480-412}{68})=P(z<1)[/tex]

[tex]P(z<1)=0.841[/tex]

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the scores of a population, and for this case we know the distribution for X is given by:

[tex]X \sim N(412,68)[/tex]  

Where [tex]\mu=412[/tex] and [tex]\sigma=68[/tex]

We are interested on this probability

[tex]P(X<480)[/tex]

And the best way to solve this problem is using the normal standard distribution and the z score given by:

[tex]z=\frac{x-\mu}{\sigma}[/tex]

If we apply this formula to our probability we got this:

[tex]P(X<480)=P(\frac{X-\mu}{\sigma}<\frac{480-\mu}{\sigma})=P(Z<\frac{480-412}{68})=P(z<1)[/tex]

And we can find this probability using the normal standard table or excel and we got:

[tex]P(z<1)=0.841[/tex]

Answer:

Paul should trim both hedges and Jared should mow both lawns.

Each neighbor would save 30 minutes per week

Step-by-step explanation:

The time each neighbor takes to finish each task is presented below:

[tex]\begin{array}{ccc}&Paul&Jared\\Mow&120&30\\Trim&90&60\end{array}[/tex]

Jared is better at mowing the lawn than trimming hedges, while Paul is better at trimming hedges than mowing the lawn. Therefore, the two neighbors could gain by specializing and trading lawn services if Paul were to trim both hedges and Jared were to mow the lawns.

The time saved by each is:

[tex]P = 210 -90-90=30\ min\\J = 90-30-30=30\ min[/tex]

Answer:

Step-by-step explanation:

given that we  have an urn with m green balls and n yellow balls. Two balls are drawn at random.

a) Assume that the balls are sampled without replacement.

m green and n yellow balls

For 2 balls to be drawn at the same colour

no of ways = either 2 green or 2 blue = mC2+nC2

Total no of ways = (m+n)C2

Prob =

= [tex]\frac{mC2 +nC2}{(m+n)C2} \\=\frac{m(m-1)+n(n-1)}{(m+n)(m+n-1)}[/tex]

=[tex]\frac{m^2+n^2-m-n}{(m+n)(m+n-1)}[/tex]

B) Assume that the balls are sampled with replacement

In this case, probability for any draw for yellow or green will be constant as

n/M+n or m/m+n respectively

Reqd prob = [tex](\frac{m}{m+n} )^2 +(\frac{n}{m+n} )^2[/tex]

=[tex]\frac{m^2+n^2}{(m+n)^2}[/tex]

c) Part B prob will be more than part a because with replacement prob is more than without replacement.

II time drawing same colour changes to m-1/.(m+n-1) if with replacement but same as m/(m+n) without replacement

[tex]\frac{m}{m+n} >\frac{m-1}{m+n-1} \\m^2+mn-m>m^2+mn-m-n\\n>0[/tex]

Since n>0 is true always, b is greater than a.

The question explores the concept of probability within scenarios of drawing balls of different colors from an urn, with and without replacement. It explores how the number of balls left in the urn changes the likelihood of drawing two balls of the same color. The answer is calculated using mathematical odds and conditions, showing that replacement affects probability especially when the total number of items (balls in this case) is small.

The subject of this question is probability, specifically conditional probability and probability with and without replacement. Here are the calculations needed to answer the question:

In a nutshell, the calculation for probability tells us how likely an outcome is, given the mathematical odds and conditions (in this case, if the balls are replaced or not after drawing).

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Answer:

The correct option is (a).

Step-by-step explanation:

The hypothesis of the study can be defined as:

H₀: The divorce rate is 50% for couples who had a child within the first two years of marriage, i.e. p = 0.50

Hₐ: The divorce rate is different from 50% for couples who had a child within the first two years of marriage, i.e. p ≠ 0.50

The 90% confidence interval is: 0.402 ± 0.056 = (0.346, 0.458) ≈ (0.35, 0.46)

The confidence level is 90%, the significance level (α) is:

[tex]\alpha =1-\frac{Confidence\ level}{100}\\=1-\frac{90}{100}\\ =0.10\ or\ 10\%[/tex]

Decision Rule:

If the null hypothesis value is not contained in the 90% confidence interval then the null hypothesis will be rejected and vice-versa.

Interpretation of the Confidence interval:

Thus all the options are correct.

The 90% confidence interval for the proportion of couples who had a child within the first two years of marriage and are divorced within five years is 0.402 ± 0.056. Based on this interval, we can conclude that the divorce rate is between 35% and 46%.

Based on the given information, the sociologist selected a random sample of 200 couples who had a child within the first two years of marriage. Out of these couples, 80 were found to be divorced within five years. The 90% confidence interval for the proportion of all couples that had a child within the first two years of marriage and are divorced within five years is given as 0.402 ± 0.056.

This means that we can be 90% confident that the true proportion of couples who had a child within the first two years of marriage and are divorced within five years lies between 0.402 - 0.056 and 0.402 + 0.056.

Therefore, the correct statement based on this interval is that the divorce rate is between 35% and 46%.

Answer:

3.945 years

Step-by-step explanation:

To answer this problem, one must know that Beethoven has composed 32 piano sonatas, 9 symphonies and 5 piano concertos.

The number of different arrangements that can be made by playing a sonata, then a symphony and then a piano concerto is:

[tex]n=32*9*5=1,440[/tex]

If a year has 365 days, the number of years that this daily policy could be continued before exactly the same program would have to be repeated is:

[tex]y=\frac{1,440}{365}=3.945\ years[/tex]

Answer:

a) E(X) = -$0.0813 , s.d (X) = 3

b) E(X) = -$0.0813 , s.d (X) = 3

c) expected loss and higher stakes of loosing.

Step-by-step explanation:

Given:

- There are total 37 slots:

           Red = 18

           Black = 18

           Green = 1

- Player on bets on either Red or black

- Wins double the bet money, loss the best is lost

Find:

a) Expected value of earnings X if we place a bet of $3

b) Expected value and standard deviation if we bet $1 each on three rounds

c) compare the two answers in a and b and comment on the riskiness of the two games

Solution:

- Define variable X as the total winnings per round. We will construct a distribution tables for total winnings per round for bets of $3 and $ 1:

- Bet: $3

         X                   -3                          3                              E(X)

        P(X)           1-0.48 = 0.5135      18/37 = 0.4864    3*(.4864-.52135) = -0.08

-The s.d(X) = sqrt(9*(0.5135 + 0.4864) - (-0.08)^2) = 3.0

- Bet: $1

         X                   -1                          1                          E (X)      

        P(X)        1-0.48 = 0.5135      18/37 = 0.4864   1*(.4864-.5135) = -0.0271

-The s.d(X) = sqrt(1*(0.5135 + 0.4864) - (-0.0271)^2) = 0.999

- The expected value for 3 rounds is:

                       E(X_1 + X_2 + X_3) = E(X_1) + E(X_2) + E(X_3)

- The above X winnings are independent from each round, hence:

                       E(3*X_1) = 3*E(X_1) = 3*(-0.0271) = -0.0813

- The standard deviation for 3 rounds is:

              sqrt(Var(X_1 + X_2 + X_3)) = sqrt(Var(X_1) + Var(X_2) + Var(X_3))

- The above X winnings are independent from each round, hence:

                       sqrt(Var(3*X_1)) = 3*Var(X_1) = 3*(0.999) = 2.9988

- For above two games are similar with an expected loss of $0.0813 for playing the game and stakes are very high due to high amount of deviation for +/- $3 of winnings.

In the game of European roulette, betting $1 in three different rounds is less risky than betting $3 in one round as it has a lower expected loss and lower variability.

In the game of roulette, the probability of winning (i.e. the ball landing on either red or black) is 18/37, and the probability of losing (the ball landing on green) is 1/37.

(a) When you bet $3 on a single round, the expected value of your winnings is given by (18/37 * $6) - (19/37 * $3) = -$0.081, and the standard deviation is sqrt([$6-$(-0.081)]^2 * 18/37 + [-$3-$(-0.081)]^2 * 19/37) = $2.899.

(b) When you bet $1 in three different rounds, the expected value of your winnings is 3 * [ (18/37 * $2) - (19/37 * $1) ] = -$0.243, and the standard deviation is sqrt(3 * [(($2-$(-0.081))^2 * 18/37) + (($-1-$(-0.081))^2 * 19/37)]) = $1.578.

(c) If we compare the two games, it is clear that betting $1 in three different rounds decreases both the expected loss and the variability of the results, indicating that the latter game is less risky.

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Answer:

When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically (0.0392 to 0.0198), to almost the half of its original value too.

Step-by-step explanation:

Poisson distribution formula

P(X=x) = f(x) = (λˣe^(-λ))/x!

λ = 0.04.

And the probability that a products one or more defects is the same thing as 1 minus the probability that a product has no defect.

P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)

P(X ≥ 1) = 1 - (0.04⁰e^(-0.04))/0! = 1 - 0.9608 = 0.0392

When the occurrence rate of defect is cut in half, that is, λ = 0.02,

P(X ≥ 1) = 1 - P(X = 0) = 1 - f(0)

P(X ≥ 1) = 1 - (0.02⁰e^(-0.02))/0! = 1 - 0.9802 = 0.0198

When the occurrence rate of defect is cut in half, the probability of a product having one or more defects drops drastically, to almost the half of its original value too.

Hope this helps!

Answer: 1

Step-by-step explanation:

If a random variable x is uniformly distributed in [a,b] the

Mean = [tex]\dfrac{a+b}{2}[/tex]

Standard deviation : [tex]\sqrt{\dfrac{(b-a)^2}{12}}[/tex]

Let x =  Times required for a cable company to fix cable problems

As per given.

x is  uniformly distributed between 40 minutes and 65 minutes.

Then , mean = [tex]\dfrac{65+40}{2}=52.5[/tex] minutes

Standard deviation : [tex]\sqrt{\dfrac{(65-40)^2}{12}}\approx7.22[/tex]minutes

Consider , P (mean- 2(Standard deviation) < X < mean+2(Standard deviation) )

= P(52.5-2(7.22)< X <  52.5+2(7.22))

=P(38.06 <X < 66.94 ).

But x lies between 40 minutes and 65 minutes.

Also,   [40 minutes,  65 minutes]⊂ [38.06 minutes , 66.94 minutes]

Therefore ,P(38.06 <X < 66.94 ) =1

∴ The probability that a randomly selected cable repair visit falls within 2 standard deviations of the mean is 1.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean,

Z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score is given by:

z = (x - μ)/σ

where x is the raw score, μ is the mean and σ is the standard deviation.

The empirical rule states that 95% of the distribution is within 2 standard deviations of the mean.

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Answer:

a) 16% of students have an SAT math score greater than 615.

b) 2.5% of students have an SAT math score greater than 715.

c) 34% of students have an SAT math score between 415 and 515.

d) [tex]Z = 1.05[/tex]

e) [tex]Z = -1.10[/tex]

Step-by-step explanation:

To solve this question, we have to understand the normal probability distribution and the empirical rule.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Empirical rule

The Empirical Rule states that, for a normally distributed random variable:

68% of the measures are within 1 standard deviation of the mean.

95% of the measures are within 2 standard deviation of the mean.

99.7% of the measures are within 3 standard deviations of the mean.

In this problem, we have that:

[tex]\mu = 515, \sigma = 100[/tex]

(a) What percentage of students have an SAT math score greater than 615?

615 is one standard deviation above the mean.

68% of the measures are within 1 standard deviation of the mean. The other 32% are more than 1 standard deviation from the mean. The normal probability distribution is symmetric. So of those 32%, 16% are more than 1 standard deviation above the mean and 16% more then 1 standard deviation below the mean.

So, 16% of students have an SAT math score greater than 615.

(b) What percentage of students have an SAT math score greater than 715?

715 is two standard deviations above the mean.

95% of the measures are within 2 standard deviations of the mean. The other 5% are more than 2 standard deviations from the mean. The normal probability distribution is symmetric. So of those 5%, 2.5% are more than 2 standard deviations above the mean and 2.5% more then 2 standard deviations below the mean.

So, 2.5% of students have an SAT math score greater than 715.

(c) What percentage of students have an SAT math score between 415 and 515?

415 is one standard deviation below the mean.

515 is the mean

68% of the measures are within 1 standard deviation of the mean. The normal probability distribution is symmetric, which means that of these 68%, 34% are within 1 standard deviation below the mean and the mean, and 34% are within the mean and 1 standard deviation above the mean.

So, 34% of students have an SAT math score between 415 and 515.

(d) What is the z-score for student with an SAT math score of 620?

We have that:

[tex]\mu = 515, \sigma = 100[/tex]

This is Z when X = 620. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{620 - 515}{100}[/tex]

[tex]Z = 1.05[/tex]

(e) What is the z-score for a student with an SAT math score of 405?

We have that:

[tex]\mu = 515, \sigma = 100[/tex]

This is Z when X = 405. So

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{405 - 515}{100}[/tex]

[tex]Z = -1.10[/tex]

a. Approximately 15.87% of students have an SAT math score greater than 615.

b. Approximately 2.28% of students have an SAT math score greater than 715.

c. 68% percentage of students have an SAT math score between 415 and 515

d. 1.05 is the z-score for student with an SAT math score of 620

e. -1.1 is the z-score for a student with an SAT math score of 405

To answer these questions, we can use the properties of a bell-shaped distribution and the empirical rule. The empirical rule states that for a bell-shaped distribution:Approximately 68% of the data falls within one standard deviation of the mean. Approximately 95% of the data falls within two standard deviations of the mean.Approximately 99.7% of the data falls within three standard deviations of the mean.

Given information: Mean (μ) = 515 and Standard Deviation (σ) = 100

(a) To find this, we need to calculate the z-score for a score of 615 and then find the percentage of data above that z-score using the standard normal distribution table.

z-score = (X - μ) / σ

z-score = (615 - 515) / 100

z-score = 1

Using the standard normal distribution table (or calculator), we find that approximately 84.13% of the data is below a z-score of 1. Since the distribution is symmetric, the percentage above the z-score of 1 is also approximately 100% - 84.13% = 15.87%.

Therefore, approximately 15.87% of students have an SAT math score greater than 615.

(b) We repeat the same process for a score of 715.

z-score = (715 - 515) / 100

z-score = 2

Using the standard normal distribution table (or calculator), we find that approximately 97.72% of the data is below a z-score of 2. The percentage above the z-score of 2 is approximately 100% - 97.72% = 2.28%.

Therefore, approximately 2.28% of students have an SAT math score greater than 715.

(c) We can use the empirical rule to find the percentage of students within one standard deviation of the mean and then subtract that from 100% to find the percentage between 415 and 515.

Percentage between 415 and 515 ≈ 68%

(d) We can calculate the z-score as follows:

z-score = (620 - 515) / 100

z-score = 1.05

(e) We can calculate the z-score as follows:

z-score = (405 - 515) / 100

z-score = -1.1

Remember that a negative z-score indicates a value below the mean.

Note: These calculations assume a normal distribution, and the actual percentages may vary slightly due to the discrete nature of test scores and rounding in calculations.

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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.

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.

Answer: 24 feet

Step-by-step explanation:

By using Pythagoras rule:

Let x be the high up the house does the ladder reached.

X^2 + 5^2= 25^2

X^2 = 25^2 - 5^2

x^2 = 625 - 25

x^2 = 600

Square both side

x = sqrt(600)

x= 24.495

x = 24 feet

Answer:

114.3

Step-by-step explanation:

If a 70kg person ingests 2L of water per day containing 8 mg/L of  dinitrotoluene, the concentration of  dinitrotoluene on that person's body is:

[tex]C=2\frac{L}{day}*8\frac{mg}{L} *\frac{1}{70\ kg}\\C=0.22857\frac{mg}{kg-day}[/tex]

The hazard ratio is defined by dividing the intake dosage (C) by the reference dose (RfD)

[tex]H=\frac{0.22857}{2*10^{-3}}\\H=114.3[/tex]

The hazard ratio is most nearly 114.3.

Answer:

(a) 0.01625

(b) 0.3483

(c)  0.15170

Step-by-step explanation:

Given that there are total of 20 study problems from which 10 will come in the exam and A student knows how to solve 15 of the problems from those 20 total problems.

(a) To calculate the probability that the student will be able to answer all 10 questions on the exam, we know that:

students will be able answer all 10 questions in the exam only when all these 10 questions will be from the 15 problem which he know how to solve.

So the chances that he knows all 10 questions in the exam = [tex]^{15}C_1_0[/tex]

And total ways in which he answer 10 question from the 20 study problems =[tex]^{20}C_1_0[/tex]

Therefore, the Probability that the student will be able to answer all 10 questions on the exam = [tex]\frac{^{15}C_1_0}{^{20}C_1_0}[/tex] = [tex]\frac{15!}{5!\times 10!} \times \frac{10!\times 10!}{20!}[/tex] {Because [tex]^{n}C_r[/tex] = [tex]\frac{n!}{r!\times (n-r)!}[/tex] }

                                                  = 0.01625

(b) Probability that the student will be able to answer exactly eight questions on the exam = In the numerator there will be No. of ways that he answer exactly eight questions from the 15 problems he knows and the remaining 2 questions he solve from the 5 questions whose answer he doesn't know and in the denominator there will Total number of ways in which he answer 10 question from the 20 study problems.

So Required Probability = [tex]\frac{^{15}C_8 \times ^{5}C_2 }{^{20}C_1_0 }[/tex]  = [tex]\frac{15!}{8!\times 7!}\times \frac{5!}{2!\times 3!}\times \frac{10!\times 10!}{20!}[/tex] = 0.3483

(c) To calculate the probability that student will be able to answer at least nine questions on the exam is given by that [He will be able to nine questions on the exam + He will be able to answer all  10 questions on the exam]

So no. of ways that he will be able to nine questions on the exam = He answer 9 questions from those 15 problems which he know and remaining one question from the other 5 questions we he don't know = [tex]^{15}C_9\times ^{5}C_1[/tex]

And no. of ways that he will be able to answer all  10 questions on the exam

= [tex]^{15}C_1_0[/tex]  

So, the required probability = [tex]\frac{(^{15}C_9\times ^{5}C_1)+^{15}C_1_0}{^{20}C_1_0}[/tex] = 0.15170

The probability that the student will be able to answer all 10 questions on the exam is 0.01625.

The probability that the student will be able to answer all 10 questions on the exam will be:

= 15C10 / 20C10

= [15! / (5! × 10!)] × [10! × 20!/20!]

= 0.01625

The probability that the student will be able to answer exactly eight questions on the exam will be:

= (15C8 × 5C2) / 20C10

= 0.3483

The probability that the student will answer at least nine questions on the exam will be:

= [(15C9 × 5C1) + 15C10] / 20C10]

= 0.15170

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The proportion of sample wafers that had at least one particle is 96%, those with at least five particles is 71%. For wafers between five and ten particles (inclusive), the proportion is 64%, and strictly between five and ten particles it is 44%. The histogram based on the given data would appear roughly bell-shaped with a right skew.

(a.) To calculate the proportion of sampled wafers that had at least one particle, we sum the frequencies of all groups with one or more particles and divide by the total sample size. So, the proportion is given by: (2+3+12+11+15+18+10+12+4+5+3+1+2+1)/100 = 96/100 = 0.96 or 96%.

Similarly, for wafers with at least five particles, the calculation is: (15+18+10+12+4+5+3+1+2+1)/100 = 71/100 = 0.71 or 71%.

(b.) For wafers with between five and ten particles (inclusive), we sum frequencies from 5 to 10 particles, giving (15+18+10+12+4+5)/100 = 64/100 = 0.64 or 64%.

For wafers strictly between five and ten particles (i.e., 6 to 9 particles inclusive), the calculation is: (18+10+12+4)/100 = 44/100 = 0.44 or 44%.

(c.) A histogram with relative frequency would show the proportion of wafers (y-axis) against the number of particles (x-axis). Given the distribution of frequencies, it might be expected that the histogram appears roughly bell-shaped, though the peak would be skewed to the right considering higher numbers around 5 to 8 particles.

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Answer:

144 observations

Step-by-step explanation:

We have that to find our [tex]\alpha[/tex] level, that is the subtraction of 1 by the confidence interval divided by 2. So:

[tex]\alpha = \frac{1-0.8684}{2} = 0.0668[/tex]

Now, we have to find z in the Ztable as such z has a pvalue of [tex]1-\alpha[/tex].

So it is z with a pvalue of [tex]1-0.0668 = 0.9332[/tex], so [tex]z = 1.5[/tex]

Now, find the margin of error M as such

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

In which [tex]\sigma[/tex] is the standard deviation of the population and n is the size of the sample.

In this problem, we have that:

[tex]\sigma = 4, M = 0.5[/tex]

We want to find n

[tex]M = z*\frac{\sigma}{\sqrt{n}}[/tex]

[tex]0.5 = 1.5*\frac{4}{\sqrt{n}}[/tex]

[tex]0.5\sqrt{n} = 6[/tex]

[tex]\sqrt{n} = \frac{6}{0.5}[/tex]

[tex]\sqrt{n} = 12[/tex]

[tex]\sqrt{n}^{2} = (12)^{2}[/tex]

[tex]n = 144[/tex]

Answer:

(a) The histogram is shown below.

(b) E (X) = 4.2

(c) SD (X) = 2.73

Step-by-step explanation:

Let X = r = a driver will tailgate the car in front of him before passing.

The probability that a driver will tailgate the car in front of him before passing is, P (X) = p = 0.35.

The sample selected is of size n = 12.

The random variable X follows a Binomial distribution with parameters n = 12 and p = 0.35.

The probability function of a binomial random variable is:

[tex]P(X=x)={n\choose x}p^{x}(1-p)^{n-x}[/tex]

(a)

For X = 0 the probability is:

[tex]P(X=0)={12\choose 0}(0.35)^{0}(1-0.35)^{12-0}=0.006[/tex]

For X = 1 the probability is:

[tex]P(X=1)={12\choose 1}(0.35)^{1}(1-0.35)^{12-1}=0.037[/tex]

For X = 2 the probability is:

[tex]P(X=2)={12\choose 2}(0.35)^{2}(1-0.35)^{12-2}=0.109[/tex]

Similarly the remaining probabilities will be computed.

The probability distribution table is shown below.

The histogram is also shown below.

(b)

The expected value of a Binomial distribution is:

[tex]E(X)=np[/tex]

The expected number of vehicles out of 12 that will tailgate is:

[tex]E(X)=np=12\times0.35=4.2[/tex]

Thus, the expected number of vehicles out of 12 that will tailgate is 4.2.

(c)

The standard deviation of a Binomial distribution is:

[tex]SD(X)=np(1-p)[/tex]

The standard deviation of vehicles out of 12 that will tailgate is:

[tex]SD(X)=np(1-p)=12\times0.35\times(1-0.35)=2.73\\[/tex]

Thus, the standard deviation of vehicles out of 12 that will tailgate is 2.73.

To determine the probability distribution, create a histogram showing the possible values of r and their probabilities. The expected number of vehicles that will tailgate can be calculated by multiplying each value of r by its probability and summing up the results. The standard deviation can be found by calculating the variance and taking the square root of it.

In order to determine the probability distribution of the number of vehicles that tailgate before passing, we need to consider the given information. We know that about 35% of all drivers tailgate. Since we are passed by 12 vehicles, the number of vehicles that tailgate can range from 0 to 12. To create a histogram showing the probability distribution, we need to calculate the probability of each possible value of r and represent them in a bar graph.

(b) To compute the expected number of vehicles that will tailgate, we need to multiply each possible value of r by its corresponding probability and sum up the results. This will give us the average number of vehicles that tailgate out of the 12 vehicles that passed us.

(c) The standard deviation of this distribution can be calculated by determining the variance and taking the square root of it. Variance is calculated by summing up the squared differences between each value of r and the expected value, multiplying each squared difference by its corresponding probability, and summing up the results.

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Answer:

a. 6

b. 18-21.9

c. 4

Step-by-step explanation:

a)

The number of classes can be assessed by counting the number of class intervals

S.no Class interval

1             10-13.9

2            14-17.9

3            18-21.9

4            22-25.9

5            26-29.9

6            30-33.9

Hence, there are 6 number of classes

b)

The class limits for third class is 18-21.9.

Upper class limit= 18

Lower class limit=21.9

c)

The class width can be calculated by taking difference of two consecutive upper class limits or two consecutive lower class limits.

Class width=14-10=4

The data represent the top speeds of players in a soccer league with 7 classes, the class limits for the third class are 18 to 21.9 km/hr, and the class width is 3.9 km/hr.

The data provided represents the top speed of soccer players in specific intervals known as classes. We can analyze the information to answer the student's queries as follows:

Based on the data provided, if we assume that the sequence goes like 10-13.9, 14-17.9, 18-21.9, and so on, we can derive the following:

Final answer:

The negations of the given predicate logic statements are expressed by switching the quantifiers (∀ and ∃) and adding negation symbols (¬) directly before the predicates, resulting in new statements that oppose the original ones.

Explanation:

The goal is to express the negation of each of the given predicate logic statements such that all negation symbols immediately precede predicates.

Answer: [tex]P(A) = 0.23[/tex]

Step-by-step explanation:

Given :

[tex]P(A/B) = 0.2[/tex]

[tex]P(A/B^{1})=0.3[/tex]

[tex]P(B)= 0.7[/tex]

[tex]P(A) = ?[/tex]

From rules of  probability :

[tex]P(A) = P(AnB) + P(A n B^{1})[/tex] ........................... equation *

[tex]P(A n B)[/tex] can be written as [tex]P(A/B)[/tex] x [tex]P(B)[/tex]

Also , [tex]P(A/B^{1})[/tex] can be written as  [tex]P(A/B^{1})[/tex] x [tex]P(B^{1})[/tex]

substituting into equation * , we have

[tex]P(A) = P(A/B)[/tex][tex]P(B) + P(A/B^{1})P(B^{1})[/tex]

since P(B) = 0.7, then [tex]P(B^{1}) = 1 - P(B) = 0.3[/tex]

so , substituting each values , we have

[tex]P(A) = (0.2)(0.7) + (0.3)(0.3)[/tex]

[tex]P(A) = 0.14 + 0.09[/tex]

[tex]P(A) = 0.23[/tex]