Chapter 7: Quiz 1
1. C
2. B
3. • Take a random sample of size n from a population.
• Compute the sample mean.
• Repeat the process of taking a random sample of size n and computing the sample mean many times.
• Plot the distribution of sample means.
4. a. I. Histogram A; n = 5
II. Histogram C; n = 10
III. Histogram B; n = 1
b. The theoretical standard error is , which turns out to be 2.597, 1.161, and 0.821 for the respective sample sizes of 1, 5, and 10. All of these are fairly close to the observed standard errors.
c. For samples of size 1, the simulated sampling distribution of the mean reflects the bimodal nature of the population distribution. For samples of sizes 5 and 10, there is a hint of skewness (that is, there is no hint of the bimodal nature of the original population), but the simulated sampling distributions appear primarily to be roughly mound-shaped.
d. The rule works moderately well for n = 5 and n = 10 and slightly less well for the bimodal distribution associated with n = 1.
5. B. Because the sample size is large (n = 100), it is reasonable to assume that the sampling distribution of the sample mean is approximately normal with mean 625 and standard deviation
Chapter 7: Quiz 2
1. C
2. D. It is equal to the sum of the variances.
3. B. For 2500 flips of a fair coin, np = 2500(0.5) = 1250 and
The reasonably likely number of heads out of 2500 flips would be approximately 1250 ± 1.96(25), or any value in the interval [1201, 1299].
4. The sampling distribution of the sample proportion for samples of size 10 is skewed to the right. The mean is 0.2, and the standard deviation is large (about 0.1265). The sampling distribution for samples of size 40 is approximately normally distributed with much less visible skewness. The mean also is 0.2, but the standard deviation is half as large (about 0.06).
5. a. You can assume randomness (and independence). Both np = 100(0.25) = 25 and n(1 – p) = 100(0.75) = 75 are at least 10, so the sampling distribution can be approximated by the normal distribution with mean and standard error
b. The z-score for 30 correct guesses is , or approximately 1.155. Thus, the probability of getting 30 or more correct guesses is approximately 0.1241. (See the plot in part a.)
c. If a person is guessing, the reasonably likely number of correct guesses is about 25 ± 1.96(4.33), or any value in the interval [16.51, 33.49]. Alternatively, the probability of getting 35 or more guesses correct is 0.0105. Either way you calculate it, 35 correct guesses is unlikely. Either the person is really lucky or she might have ESP.
6. a. You were told that the two trains were selected randomly and independently, so the mean and standard error of the sampling distribution of the sum are
The sampling distribution of the sum will be approximately normal because the original population of late times is normal. The z-score is The probability that the sum of their minutes late is less than 10 is about 0.017.
b. The mean and standard error of the sampling distribution of the difference is
The sampling distribution of the difference will be approximately normal because the original population of late times is normal. The z-score is The probability that the second train selected is at least 2 minutes later than the first train is about 0.2398.