Section 5.2 Solutions

 

5.33

a.      There is variability in the sampling process. That is, different random samples produce different sample means. Some of the sample mean values will be smaller than  and some of them will be larger than , but the average of all the possible sample means (from all possible samples) will be exactly . This is what it means for  to be an unbiased estimator of .

 

b.      A larger sample provides more information—more observations on which to base the sample mean. Hence, sample mean values based on larger samples are more precise (less variability), so they are closer to their mean, .

 

 

5.36

a.      The mean is , and the standard deviation is

 

b.      and

 

 

c.       From the probability distribution given, . Using the Central Limit Theorem (since n = 50 is large),  (Although a normal curve picture isn’t included with this solution, I encourage you draw one.)

 

 

5.37

(Although normal curve pictures aren’t included with these solutions, I encourage you draw them.)

 

The population of individual glucose measurements is normal with  and  Let the mean of 4 glucose levels. Then  has a normal distribution with and  (This holds, even for the small sample, since the individual measurements are normally distributed.)

 

a.      Based on a single measurement, the probability Sheila is diagnosed as having gestational diabetes is .

 

b.      Based on the average of four measurements, the probability Sheila is diagnosed as having gestational diabetes is .

 

 

5.39

(Although a normal curve picture isn’t included with this solution, I encourage you draw it.)

 

The population of individual glucose measurements is normal with  and  Let the mean of 4 glucose levels. Then  has a normal distribution with and

 

We want to find L such that  We know that the 95th percentile of the standard normal distribution is z = 1.645 (you can also use 1.64 or 1.65). Thus,  The level is 133.225 mg/dl.

 

 

5.41

(Although a normal curve picture isn’t included with this solution, I encourage you draw it.)

 

The population of passenger weights has mean  pounds and standard deviation  pounds. The population is not normal, but it’s not very non-normal. Hence, a sample size of 19 is probably large enough to use the Central Limit Theorem (even though our quick check of  isn’t met).

 

Then we know the total weight of 19 passengers, , is approximately normal with mean  pounds and standard deviation  pounds.

 

Then,

. So the approximate probability is 0.0052 that the total weight of the passengers exceeds 4000 pounds.

 

Alternative method (using the sampling distribution of ):

 

 

5.44

(Although normal curve pictures aren’t included with these solutions, I encourage you draw them.)

 

Let X be the diameter of a randomly selected shaft. Then X is N(2.45, 0.01).

 

a.      Then the probability a randomly selected shaft fits into the hole is . Hence, 99.38% of the shafts fit into the hole.

 

b.      Let Y be the hole diameters. Then Y is N(2.5, 0.01). Furthermore, X and Y are independent. We need to find P(Y > X + 0.025) = P(Y – X > 0.025). We know that Y – X has a normal distribution with mean  and standard deviation .

 

Hence,

 

 

5.45

(Although normal curve pictures aren’t included with these solutions, I encourage you draw them.)

 

Let X be the breaking strength of untreated fabric. Then X is N(58, 2.3). Also, let Y be the breaking strength of treated fabric. Then Y is N(30, 1.6). Let  be the mean breaking strength of 5 untreated specimens. Then  is . Let  be the mean breaking strength of 5 treated specimens. Then  is . Furthermore,  and  are independent since they are based on independent measurements. Then  also has a normal distribution with mean  and standard deviation .

 

a.     

 

b.     

 

 

5.46

(Although a normal curve picture isn’t included with this solution, I encourage you draw it.)

 

Let Y be a randomly selected Journal score. Then Y is N(4.8, 1.5). Also, let X be a randomly selected Enquirer score. Then X is N(2.4, 1.6).

 

a.      The individual scores are discrete (1, 2, 3, 4, 5, 6, or 7), but the average of 30 scores can be anything between 1 and 7 (for example, the mean can be 4.5). Since the sample size is “large” (n = 30), the Central Limit Theorem tells us that the sample mean score will be approximately normal.

 

b.      Let  be the mean Enquirer score for 30 students. Then  is approximately . Let  be the mean Journal score for 30 students. Then  is .

 

 

c.       We know that  and  are independent. Then  has an approximate normal distribution with mean  and standard deviation .

 

d.      Then,

 

 

 

5.50

(Although a normal curve picture isn’t included with this solution, I encourage you draw it.)

 

The distribution of household monthly fees for service providers has mean $28 and standard deviation $10. Let  be the sample average fee for a random sample of 500 households. Since n = 500 is large, the Central Limit Theorem tells us that  has an approximate normal distribution with mean  and standard deviation .

 

Then .