Q: A pollster wishes to estimate the number of left-handed scientists. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 2%? A previous study indicated that the proportion of left-handed scientists is 9%.

$$n = \left(\frac{z_{0.05} \sqrt{0.09 \times 0.91}}{0.02}\$$right)^2$$ \approx 300$$

Q: The lengths of human pregnancies are approximately normally distributed with a mean of 266 days and a standard deviation of 16 days. Find the 70th percentile for the length of pregnancies.

$$266 + (0.524 \times 16) \approx 274.4$$ days

Q: The lengths of human pregnancies are approximately normally distributed with a mean of 266 days and a standard deviation of 16 days. Find the minimum and maximum values for the middle 20% of pregnancies.

Minimum: $$266 + (-0.253 \times 16) \approx 262.0$$ days; Maximum: $$266 + (0.253 \times 16) \approx 270.0$$ days

Question 1

When is the sampling distribution of all sample means normally distributed?

Accepted Answer

Either a) or b)

Question 2

If the sample size is increased, how is the margin of error for a confidence interval affected, everything else being equal?

Accepted Answer

It is decreased.

Question 3

Refer to the normal curve shown. If one individual value is selected at random, what is the probability that it is greater than 849?

Accepted Answer

0.5

Question 4

A random sample of 1493 adults was asked, 'Are you afraid to go out at night within a mile of your home because of crime?' Of that sample, 672 say 'Yes.' What is the point estimate $$p$$ for the proportion of all adults who would say they are afraid? Round to the nearest thousandth.

Accepted Answer

$$p = \frac{672}{1493} \approx 0.451$$

Question 5

For the sample in the previous question, does the sampling distribution of sample proportions have a normal distribution? Justify your answer.

Accepted Answer

Yes, because $$np \geq 5$$ and $$n(1-p) \geq 5.$$

Question 6

A newspaper reported that 21% of all adults in the U.S. are illiterate. Describe the sampling distribution of sample proportions for samples of 125 adults.

Accepted Answer

Approximately normal with mean $$\mu_p = 0.21$$ and standard deviation $$\sigma_p = \sqrt{\frac{0.21 	imes 0.79}{125}}$$

Question 7

The repair costs for microwave ovens is normally distributed with a mean of $$57 and a standard deviation of 8. $$For a random sample of 12 ovens, what is the probability that the average cost is more than $60?

Accepted Answer

$$P(\bar{x} \u003e 60) = P\left(z \u003e \frac{60-57}{\frac{8}{\sqrt{12}}}\right) \approx P(z \u003e 1.30) \approx 0.0968$$

Question 8

A pollster wishes to estimate the number of left-handed scientists. How large a sample is needed in order to be 90% confident that the sample proportion will not differ from the true proportion by more than 2%? A previous study indicated that the proportion of left-handed scientists is 9%.

Accepted Answer

$$n = \left(\frac{z_{0.05} \sqrt{0.09 	imes 0.91}}{0.02}\$$right)^2$$ \approx 300$$

Question 9

In a large class, the final exam grades are normally distributed with a mean of 72 and a standard deviation of 13. What is the probability that a randomly chosen student gets a grade above 75?

Accepted Answer

$$P(X \u003e 75) = P\left(z \u003e \frac{75-72}{13}\right) = P(z \u003e 0.23) \approx 0.4090$$

Question 10

In a large class, the final exam grades are normally distributed with a mean of 72 and a standard deviation of 13. What is the probability that the grade is between 70 and 80?

Accepted Answer

$$P(70 \u003c X \u003c 80) = P\left(\frac{70-72}{13} \u003c z \u003c \frac{80-72}{13}\right) = P(-0.15 \u003c z \u003c 0.62) \approx 0.2876$$

Question 11

A study looked at the amount spent on supermarket impulse buying in a 10-minute (unplanned) shopping interval. The mean amount spent is $$21.23 and the standard deviation is 7.86. $$The shape of the distribution was right-skewed. Describe the sampling distribution of sample means based on samples of 70 impulse shoppers each.

Accepted Answer

Approximately normal with mean $$\mu = 21.23$$ and standard deviation $$\sigma = \frac{7.86}{\sqrt{70}}$$

Question 12

The lengths of human pregnancies are approximately normally distributed with a mean of 266 days and a standard deviation of 16 days. Find the 70th percentile for the length of pregnancies.

Accepted Answer

$$266 + (0.524 	imes 16) \approx 274.4$$ days

Question 13

The lengths of human pregnancies are approximately normally distributed with a mean of 266 days and a standard deviation of 16 days. Find the minimum and maximum values for the middle 20% of pregnancies.

Accepted Answer

Minimum: $$266 + (-0.253 	imes 16) \approx 262.0$$ days; Maximum: $$266 + (0.253 	imes 16) \approx 270.0$$ days

Question 14

A city builds a new public parking garage and wants to estimate the mean amount collected daily. From a random sample of 43 days, the sample mean is $$126 and the sample standard deviation is 15. $$Calculate the 95% confidence interval for the mean daily amount collected overall.

Accepted Answer

$$126 \pm t_{0.025,42} 	imes \frac{15}{\sqrt{43}} \approx [121.43, 130.57]$$

Question 15

A city builds a new public parking garage and wants to estimate the mean amount collected daily. From a random sample of 43 days, the sample mean is $$126 and the sample standard deviation is 15. $$Interpret the 95% confidence interval you calculated.

Accepted Answer

We are 95% confident that the true mean daily amount collected is between $$121.43 and 130.57.$$

Practice Exam Guidance: Sampling Distributions, Confidence Intervals, and Normal Probability

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