Distribution Lab Quiz
Distribution Lab Quiz
Topic: Central Limit Theorem & Sampling Distributions
Scientist:
Date:
1
The heights of 18-year-old men are normally distributed with mean \( \mu = 68 \) inches and standard deviation \( \sigma = 3 \) inches. If one man is selected at random, what is the probability that he is between 67 and 69 inches tall?
Table Look-up
| z | P(Z ≤ z) |
|---|
| 0.33 | 0.6293 |
| -0.33 | 0.3707 |
Calculation Zone
0.2586
0.6293
0.3413
0.5000
2
Using the same distribution (\( \mu = 68, \sigma = 3 \)), if a random sample of \( n = 9 \) men is selected, what is the probability that the sample mean height \( \bar{x} \) is between 67 and 69 inches?
Table Look-up
| z | P(Z ≤ z) |
|---|
| 1.00 | 0.8413 |
| -1.00 | 0.1587 |
Calculation Zone
0.3413
0.6826
0.9544
0.4772
3
Blood glucose levels are normally distributed with \( \mu = 85 \) and \( \sigma = 25 \). If a doctor takes the average of \( n = 2 \) tests, what is the standard error \( \sigma_{\bar{x}} \) of the sampling distribution?
Calculation Zone
25.00
17.68
12.50
5.59
4
White blood cell counts are approximately normal with \( \mu = 7500 \) and \( \sigma = 1750 \). What is the probability that a single test result \( x \) is less than 3500?
Table Look-up
| z | P(Z ≤ z) |
|---|
| -2.29 | 0.0110 |
| -1.29 | 0.0985 |
Calculation Zone
0.0110
0.9890
0.0985
0.1750
5
Deer weights are approximately normal with \( \mu = 63.0 \) kg and \( \sigma = 7.1 \) kg. What is the probability that a single captured doe weighs less than 54 kg?
Table Look-up
| z | P(Z ≤ z) |
|---|
| -1.27 | 0.1020 |
| -0.27 | 0.3936 |
Calculation Zone
0.8980
0.1020
0.3936
0.0510
6
Incubation times have \( \mu = 16 \) days and \( \sigma = 2 \) days. For \( n = 30 \) eggs, the standard error \( \sigma_{\bar{x}} \) is calculated as:
Calculation Zone
2.000 days
0.365 days
0.067 days
1.095 days
7
Monthly returns have \( \mu = 1.6\% \) and \( \sigma = 0.9\% \). For a 24-month period, what is the probability that the average monthly return \( \bar{x} \) is between 1% and 2%?
Table Look-up
| z | P(Z ≤ z) |
|---|
| 2.18 | 0.9854 |
| -3.27 | 0.0005 |
Calculation Zone
0.9849
0.9995
0.0145
0.5000
8
According to the Central Limit Theorem, what happens to the standard deviation of the sampling distribution of the mean (\( \sigma_{\bar{x}} \)) as the sample size \( n \) increases?
It increases proportionally to \( \sqrt{n} \)
It decreases as the sample size gets larger
It remains constant and equal to \( \sigma \)
It becomes equal to the population mean \( \mu \)
Distribution Lab Answer Key
Distribution Lab Answer Key
Teacher Reference & Solution Guide
Master Key
1
Answer: 0.2586
\( z = (x - \mu)/\sigma \)
\( z_1 = -0.33 \rightarrow 0.3707 \)
\( z_2 = 0.33 \rightarrow 0.6293 \)
Area = \( 0.6293 - 0.3707 \)
2
Answer: 0.6826
\( \sigma_{\bar{x}} = \sigma / \sqrt{n} = 3 / \sqrt{9} = 1 \)
\( z_1 = -1.00 \rightarrow 0.1587 \)
\( z_2 = 1.00 \rightarrow 0.8413 \)
Area = \( 0.8413 - 0.1587 \)
3
Answer: 17.68
Standard Error calculation:
\( \sigma_{\bar{x}} = 25 / \sqrt{2} \approx 17.68 \)
Sample mean distribution is normal since population is normal.
4
Answer: 0.0110
\( z = (3500 - 7500) / 1750 = -2.29 \)
Look-up: \( P(Z \le -2.29) = 0.0110 \)
5
Answer: 0.1020
\( z = (54 - 63) / 7.1 = -1.27 \)
Look-up: \( P(Z \le -1.27) = 0.1020 \)
6
Answer: 0.365 days
\( \sigma_{\bar{x}} = 2 / \sqrt{30} \approx 0.365 \)
Note: \( n=30 \) satisfies CLT requirement for normality regardless of population shape.
7
Answer: 0.9849
\( \sigma_{\bar{x}} = 0.9 / \sqrt{24} = 0.1837 \)
\( z_1 = -3.27, z_2 = 2.18 \)
\( P = 0.9854 - 0.0005 = 0.9849 \)
8
Answer: It decreases
Central Limit Theorem states that as \( n \uparrow \), variability \( \sigma_{\bar{x}} \downarrow \).
The distribution also becomes more normal.
Z-Table Master Reference
z = 0.33
0.6293
z = 1.00
0.8413
z = -2.29
0.0110
z = -1.27
0.1020
