Math 221 **** Example Format ****
Week 6 Lab

Submitted by: (Insert Name)

Part 1. Normal Distributions and Birth Weights in America

1(a) 37 to 39 weeks as mean is around 7.33 lb.
1(b) 40 weeks as mean is around 7.72 lb.
1(c) 28 to 31 weeks as mean is 4.07 lb.

2(a) 99.88%, Excel command used was NORMDIST(5.5,1.88,1.19,TRUE).
2(b) 43.83%
2(c) 4.66%
2(d) 2.75%

3(a) Above 8.7269, Excel command used was NORMINV(0.9,7.33,1.09).
3(b) Above 9.0853

4(a) 41.41%, Excel command used was NORMDIST(9,5.73,1.48,TRUE)-NORMDIST(6,5.73,1.48,TRUE).
4(b) 82.61%
4(c) 81.56%

5(a) 88.36%, Excel command used was NORMDIST(3.3,1.88,1.19,TRUE)
5(b) 5.03%
5(c) 0.01%

Part 2. Age Distribution in the United States

1. 36.48
2. 36.21, Excel command used Average(). It is almost the same as the mean age in the United States. It agrees with the central limit theorem as it states that population mean should be equal to sample mean.
3. No, because histogram is not bell shaped.
4.
Maximum value = 44.72
Minimum value = 28.14
Class width = (44.72-28.14)/9 = 1.85(after rounding up)

Class Frequency relative frequency
28.14 - 29.99 1 2.8%
29.99 - 31.84 3 8.3%
31.84 - 33.69 3 8.3%
33.69 - 35.54 7 19.4%
35.54 - 37.39 11 30.6%
37.39 - 39.24 5 13.9%
39.24 - 41.09 3 8.3%
41.09 - 42.94 2 5.6%
42.94 - 44.79 1 2.8% 36 1

We can note that histogram is bell shaped and symmetric. It agrees with the results predicted by the Central Limit Theorem which states that for “large” samples, the sampling distribution is approximately normal.

5. Standard deviation of the set of 36 sample means = 3.551804

Population standard deviation = 22.57

Sample standard deviation = = 3.57

Since both the values are very close so it agrees with the result predicted by the Central Limit Theorem.

Part 3. Finding z- and t-scores for Confidence Intervals

1. Using Excel, find the z-score that corresponds to the following Confidence Levels:

a. 80%
Answer: 1.282

b. 85%
Answer: 1.44

c. 92%
Answer: 1.751

d. 97%
Answer: 2.17

2. Using Excel, find the t-score that corresponds to the following Confidence Levels and Sample Sizes:

a. 95% with n = 25
Answer: 2.064

b. 96% with n = 15
Answer: 2.264

c. 97% with n = 21
Answer: 2.336

d. 91% with n = 10
Answer: 1.899

3. Suppose we wish to estimate the population mean using a confidence interval. When is it appropriate to use a z-score? When is it appropriate to use a t-score?

When our sample size is 30 or larger then distribution of sample means is normal then we can use z-score.

When our sample size is less than 30 and we assume that the population is normal or approximately normal, if population standard deviation is unknown, we should use t-score but if population standard deviation is known we can use z-score.

Part 4. Bob’s Candies

1. Find the sample mean and sample standard deviation of the amount citizens spend per year.
Sample mean = 78.4
Sample standard deviation = 6.21

2. When finding a confidence interval for the true mean spent of ALL citizens, should we use a z-score or a t-score? Why?

We will use z-score as sample size is larger than 30.

3. Find the z/t-values (as appropriate) for a 95% confidence interval and a 92% confidence interval. z-values for a 95% confidence interval = 1.96 z-values for a 92% confidence interval = 1.751

4. Find a 95% and a 92% confidence interval for the true mean amount that citizens spend per year.

95% confidence interval for the true mean amount that citizens spend per year

95% confidence interval for the true mean amount that citizens spend per year is (76.5, 80.3).

92% confidence interval for the true mean amount that citizens spend per year

92% confidence interval for the true mean amount that citizens spend per year is (76.7, 80.1).

5. What do you think the lowest possible mean amount spent per year is? Why?

Based on the above confidence intervals lowest possible mean amount spent per year is 76.5. We can say that lowest possible mean amount spent per year will be greater than $75.

6. Do you think Bob has a good customer base for his new business? Explain.

Since the citizens in his area spend more than $75 per year so we can say that Bob has a good customer base for his new business.