Math 207 – Solutions to Assignment 5

 

6.25

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

 

The basal diameter, X, of a sunflower plant is normally distributed with  = 35 mm and  = 3 mm.

 

a.     

 

b.      The two selections are made without replacement, so technically the selections are not independent. By our rough rule, though (since the population of sunflowers is probably larger than 20 times the sample size of 2) the selections are approximately independent. Then P(first diameter greater than 40  second diameter greater than 20) = P(first diameter greater than 40)P(second diameter greater than 40) = (0.0475)(0.0475) = 0.0023.

 

c.       Using the Empirical Rule, we expect about 95% of the observations to be within 2 standard deviations of the mean. Hence, we would expect 95% of the diameters to lie within the values (35 – 2(3), 35 + 2(3)) = (29 mm, 41 mm).

 

Using the normal distribution exactly, we expect 95% of the observations to be within 1.96 standard deviations of the mean. Hence, we would expect 95% of the diameters to lie within the values (35 – 1.96(3), 35 + 1.96(3)) = (29.12 mm, 40.88 mm).

 

d.      We want to find x such that

 

We know that the 90th percentile of the standard normal distribution is 1.28 (using Table 3; you could also use 1.29, or interpolate between 1.28 and 1.29). That is, x is 1.28 standard deviations above the mean. Therefore,

So, 38.84 mm is the 90th percentile of the distribution of diameters.

 

 

6.30

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

 

The grain amount, X, discharged from a grain loader is normally distributed with mean  and  = 25.7 bushels. The loader fills containers that can hold 2000 bushels of grain. At what value should  be set, so that there is only a 0.01 chance of a container being overfilled (i.e., filled with more than 2000 bushels)?

 

We want to find  such that

 

 

We know the 99th percentile of the standard normal distribution is 2.33 (using Table 3; you could also use 2.32, or interpolate between 2.32 and 2.33). Therefore,  So the mean should be set at about 1940 bushels.

 

6.47

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

 

Let X be the number of people who made reservations and actually show up. Then X has a binomial distribution with n = 215 and p = 0.9 (assuming that people decide to show up independently of each other).

 

Is the normal approximation appropriate here? Yes, since np = 215(0.9) = 193.5 > 5 and n(1 – p) = 215(0.1) = 21.5 > 5.

 

Then X is approximately normal with  and

 

In order for everyone to have a room, 200 or fewer people will have to show up, and this probability is

 

Additional Solution Methods:

·         Let X be the number of people who don’t show up. Then X is approximately normal with  and , and we’re looking for

 

 

·         Also this problem could be rephrased in terms of the proportion of people who show up. We would then need to find where  is approximately normal with and  The answer will be exactly the same (with the possible exception of rounding).

 

 

7.6

The questionnaire was sent to a random sample of voters, yet the responses weren’t from a random sample. This is a voluntary response sample, where people who have strong feelings (especially negative) about the issue are more likely to respond. Therefore, the sample is most likely biased, and the 72% figure is probably too high an estimate. In order for the sample not to be biased, the people conducting the study would have to work hard (e.g., many follow-up contacts) to get information from all 1000 people in the random sample.

 

 

7.30

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

 

  1. Since the individuals strength measurements are normally distributed, the sampling distribution of  is exactly normal (regardless of sample size) with and .
  3. We need to find  such that . We know the 0.1st percentile of the standard normal distribution is –3.08 (from Table 3; you could also use -3.09 or -3.10). Therefore, . So the mean should be set at 21.95 pounds per square inch.

 

7.70

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

 

Since the population size of Americans (if this is the population of interest) is obviously larger than 20 times the sample size of 500, we have approximate independence. Hence, this is a binomial setting with n = 500 and p = 0.85.

 

a.       and

 

b.      Since np = 500(0.85) = 425 > 5 and n(1 – p) = 500(1 – 0.85) = 75 > 5, the normal approximation should be good. Hence,  has an approximate normal distribution.

 

c.      

 

 

d.     

 

e.      How many standard errors do we need to move up and down from the mean in order encompass 99% of the sample proportion values? To answer this, we need to determine the 99.5th percentile from the standard normal table. That value is 2.575 (or you could use 2.57 or 2.58). Hence, 99% of the time, the sample proportion will like between the limits    (0.81, 0.89).

 

 

 

7.72

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

 

Because the individual weights are mound-shaped, even for a small sample size the distribution of the total weight,  is approximately normal with mean 150n and standard deviation . Then we want to find n such that .

 

We know that the 99th percentile of the standard normal distribution is about 2.33 (from Table 3; you could also use 2.32, or interpolate between 2.32 and 2.33). So we want to find n such that,

 

 

Now,  for

 

For n = 15, the condition that the total weight exceeds 2000 with small probability is not met:  and total weight exceeds 2000 with probability 0.9678 (this is obviously much bigger than 0.01).

 

For n = 11,  and the total weight exceeds 2000 with probability 0.0013 (this is less than 0.01).

 

For n = 12,  and the total weight exceeds 2000 with probability 0.0495 (this is bigger than 0.01).

 

Hence, the largest number of people is 11.

 

 

7.81

Note: Although normal curve pictures are not included with this solution (simply because of the limitations of Word), I strongly encourage you to draw appropriate pictures as part of your solution.

a.      Since the individual fill amounts follow a normal distribution, the distribution of the total fill,  for a case of 24 cans is exactly normal with mean  fluid ounces and standard deviation  fluid ounces.

 

b.     

 

c.       The average fill of a 6-pack, , has an exact normal distribution with mean 12 fluid ounces and standard deviation fluid ounces. Then,