Chapter 7, Problem 31CP

Practicing the Basics

Exam performance An exam consists of 50 multiple- choice questions. Based on how much you studied, for any given question you think you have a probability of p = 0.70 of getting the correct answer. Consider the sampling distribution of the sample proportion of the 50 questions on which you get the correct answer.

1. a. Find the mean and standard deviation of the sampling distribution of this proportion.
2. b. What do you expect for the shape of the sampling distribution? Why?
3. c. If truly p = 0.70, would it be very surprising if you got correct answers on only 60% of the questions? Justify your answer by using the normal distribution to approximate the probability of a sample proportion of 0.60 or less.

To determine

Identify the mean and standard deviation of the sampling distribution for the sample of size 50.

Answer to Problem 31CP

The mean (p) of the sampling distribution is 0.70.

The standard deviation of the sampling distribution is 0.0648.

Explanation of Solution

Calculation:

It is given that the probability of getting correct answers for the sample of 50 multiple choice questions is 0.70.

Denote p as the sample proportion and it is given as 0.70.

Mean of the sampling distribution:

The mean of the distribution of the sample proportion will be equals to the proportion (p).

Thus, the mean of the distribution is 0.70.

Standard deviation of the sampling distribution:

$\text{Standard deviation}=\sqrt{\frac{p\left(1-p\right)}{n}}$.

Where, p be the sample proportion and n be the sample size.

Substitute the corresponding values to get standard deviation,

$\begin{array}{c}\text{Standard deviation}=\sqrt{\frac{0.70\left(1-0.70\right)}{50}}\\ =\sqrt{\frac{0.70\left(1-0.70\right)}{50}}\\ =\sqrt{\frac{0.70\left(0.30\right)}{50}}\end{array}$

                           $\begin{array}{c}=\sqrt{\frac{0.21}{50}}\\ =\sqrt{0.0042}\\ \approx 0.0648\end{array}$

Thus, the standard deviation of the sampling distribution is 0.0648.

To determine

Identify the shape of the distribution.

Answer to Problem 31CP

The shape of the distribution is said to be symmetric or bell-shaped.

Explanation of Solution

Calculation:

Verify the condition:

For the large sample size n, the binomial distribution is approximately normal, only if the following conditions are satisfied. That is, $np\ge 15$ and $n\left(1-p\right)\ge 15$.

Requirement check:

Substitute the corresponding values to check the requirements,

$\begin{array}{c}np=50\left(0.70\right)\\ \approx 35\\ \ge 15\end{array}$

$\begin{array}{c}n\left(1-p\right)=50\left(1-0.7\right)\\ =50\left(0.30\right)\\ \approx 15\\ \ge 15\end{array}$

Here, the conditions for the normality assumption have been satisfied. Thus, the shape of the distribution is assumed to be symmetric or bell-shaped.

To determine

Explain whether it would be surprising of the proportion of getting correct answers are only 60%.

Answer to Problem 31CP

No, it would not be surprising of the proportion of getting correct answers are only 60%.

Explanation of Solution

Calculation:

The z-score:

The z-score of an observation is obtained by subtracting the mean value from the observation and dividing the difference by the corresponding standard deviation. It measures the number of standard deviations that an observation is away from its mean. A positive z-score indicates that the observation lies above the mean, whereas a negative z-score indicates that the observation lies below the mean. The formula for the z-score is:

$z=\frac{\text{observation}-\text{mean}}{\text{standard deviation}}$.

Here, $\text{mean}\left(p\right)=0.60$, $\text{standard deviation}=0.0648$.

For this sample proportion, $\text{observation}=0.60$.

Thus, the z-score for sample proportion is:

$\begin{array}{c}z=\frac{\text{observation}-\text{mean}}{\text{standard deviation}}\\ =\frac{0.60-0.70}{0.0648}\\ =\frac{-0.10}{0.0648}\\ \approx -1.54\end{array}$

Thus, the z-score for sample proportion of 0.60 is –1.54.

Use Table A in Appendix A: Standard Normal Cumulative Probabilities to find the z value.

Procedure:

For z at –1.54,

• Locate 1.50 in the left column of the table.
• Obtain the value in the corresponding row below 0.04.

That is, $P\left(z<-1.54\right)=0.063$

Thus, the probability of the sample proportion almost 0.60 is 0.063.

However, the probability of 0.063 is very low, so, it would not be surprising if the proportion of getting correct answers is only 60%.