The probability that the sample proportion lies between 0.59 and 0.65.

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.

Answer 6

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

Answer 7

Chapter 5.3, Problem 71E

(a)

To determine

To find: The probability that the sample proportion lies between 0.59 and 0.65.

Answer 8

(a)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.59 and 0.65 is 0.832.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.62=310

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.62×0.38=10.85

Also,

n(1−p)=500×(1−0.62)=190

As np and n(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 310 and variance 117.8.

Now, obtain the Z-scores for p1 and p2 as shown below,

The Z score for p1=0.59 is calculated as,

Z−=np1−npnpq=500×0.59−31010.85=−1.38

The Z score for p2=0.65 is calculated as,

Z+=np2−npnpq=500×0.65−31010.85=1.38

Now, calculate the probability that the sample proportion p^ of those who received nasty messages falls between 0.59 and 0.65 as shown below,

P(0.59≤p^≤0.65)=P(−1.38≤Z≤1.38)=P(Z≤1.38)−P(Z≤−1.38)

By using the standard normal table, the areas corresponding to probabilities P(Z≤1.38) and P(Z≤−1.38) are 0.916 and 0.084, respectively.

So,

P(0.59≤p^≤0.65)=P(Z≤1.38)−P(Z≤−1.38)=0.916−0.084=0.832

Hence, the required probability is 0.832.

Answer 13

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

Answer 14

(b)

To determine

To find: The probability that the sample proportion lies between 0.87 and 0.93.

Answer 15

(b)

Expert Solution

Answer to Problem 71E

Solution: The probability that a randomly selected sample has proportion between 0.87 and 0.93 is 0.974.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation: The mean of the binomial distribution is calculated as.

n×p=500×0.90=450

The standard deviation (σX) of the binomial distribution is calculated as,

σX=np(1−p)=500×0.9×0.1=6.7

Also,

np(1−p)=500×0.9×(1−0.9)=45

As np and np(1−p) are greater than 10, the binomial distribution can be approximated to normal distribution with mean 450 and standard deviation 6.7.

Now, obtain the Z scores for p1 and p2 as shown below,

The Z score for p1=0.87 is calculated as,

Z−=np1−npnpq=500×0.87−4506.7=−2.23

The Z score for p2=0.93 is calculated as,

Z+=np2−npnpq=500×0.93−4506.7=2.23

Now, calculate the probability that the sample proportion p^ falls between 0.87 and 0.93 as shown below,

P(0.87≤p^≤0.93)=P(−2.23≤Z≤2.23)=P(Z≤2.23)−P(Z≤−2.23)

By using the standard normal table, the areas corresponding to probabilities P(Z≤2.23) and P(Z≤−2.23) are 0.987 and 0.012 respectively.

So,

P(0.87≤p^≤0.93)=P(Z≤2.23)−P(Z≤−2.23)=0.987−0.012=0.975

Hence, the required probability is 0.975.

Answer 20

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.

Answer 21

(c)

To determine

To explain: The effect on the probability of sample proportion p^ - that it falls within ±3% of the population proportion (p) value as the value of p gets closer to 1, by using the results of parts (a) and (b).

Answer 22

(c)

Expert Solution

Answer to Problem 71E

Solution: As the value of true population proportion p goes closer to 1, the probability of sample proportion p^ falling within ±3% will increase.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

The value of probability depends on the standard deviation. When true proportion p goes closer to one, it will attain the maximum value and then the standard deviation will decreases, so the overall value of probability will increase.