Starting with a population that is normally distributed with a mean of 100 and a standard deviation of 12, answer the following questions (if possible).

a. What is the probability of randomly drawing a single score between 100 and 106, p(100 < X < 106)?
b. What is the probability of drawing a random sample of 9 observations with a mean between 100 and 106, p(100 < M < 106)?
c. What is the probability of drawing a random sample of 36 observations with a mean between 100 and 106, p(100 < M < 106)?

a. Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportions related to the Z scores calculated.

b, c. Z = (score-mean)/SEm

SEm = SD/√n

Use the same table.

To answer these questions, we will use the properties of the normal distribution. The normal distribution is characterized by its mean (µ) and standard deviation (σ). In this case, the population has a mean of 100 and a standard deviation of 12.

a. To find the probability of randomly drawing a single score between 100 and 106, p(100 < X < 106), we can use the Z-score formula. The Z-score formula calculates the number of standard deviations a particular value is from the mean.

The Z-score is calculated as: Z = (X - µ) / σ

In this case, we want to find the probability between two scores, so we need to calculate the Z-scores for both 100 and 106.

Z1 = (100 - 100) / 12 = 0
Z2 = (106 - 100) / 12 = 0.5

Using a Z-table or a statistical calculator, we can find the probability associated with each Z-score. The probability associated with Z1 (0) is 0.5000, and the probability associated with Z2 (0.5) is 0.6915.

To find the probability between these two scores, we subtract the probability associated with Z1 from the probability associated with Z2:

p(100 < X < 106) = 0.6915 - 0.5000 = 0.1915

Therefore, the probability of randomly drawing a single score between 100 and 106 is approximately 0.1915.

b. To find the probability of drawing a random sample of 9 observations with a mean between 100 and 106, p(100 < M < 106), we need to use the Central Limit Theorem. According to the Central Limit Theorem, the distribution of sample means approaches a normal distribution with a mean equal to the population mean (µ) and a standard deviation equal to the population standard deviation (σ) divided by the square root of the sample size (√n).

In this case, the sample size is 9, so we can calculate the standard deviation of the sample mean (σM) as σ / √n:

σM = 12 / √9 = 12 / 3 = 4

Now, we can calculate the Z-scores for both 100 and 106, using the formula Z = (X - µ) / σM:

Z1 = (100 - 100) / 4 = 0
Z2 = (106 - 100) / 4 = 1.5

Using a Z-table or a statistical calculator, we can find the probability associated with each Z-score. The probability associated with Z1 (0) is 0.5000, and the probability associated with Z2 (1.5) is 0.9332.

To find the probability between these two means, we subtract the probability associated with Z1 from the probability associated with Z2:

p(100 < M < 106) = 0.9332 - 0.5000 = 0.4332

Therefore, the probability of drawing a random sample of 9 observations with a mean between 100 and 106 is approximately 0.4332.

c. To find the probability of drawing a random sample of 36 observations with a mean between 100 and 106, p(100 < M < 106), we can use the same approach as in part b, but with the adjusted sample size.

The standard deviation of the sample mean (σM) is calculated as σ / √n:

σM = 12 / √36 = 12 / 6 = 2

Now, we can calculate the Z-scores for both 100 and 106, using the formula Z = (X - µ) / σM:

Z1 = (100 - 100) / 2 = 0
Z2 = (106 - 100) / 2 = 3

Using a Z-table or a statistical calculator, we can find the probability associated with each Z-score. The probability associated with Z1 (0) is 0.5000, and the probability associated with Z2 (3) is 0.9987.

To find the probability between these two means, we subtract the probability associated with Z1 from the probability associated with Z2:

p(100 < M < 106) = 0.9987 - 0.5000 = 0.4987

Therefore, the probability of drawing a random sample of 36 observations with a mean between 100 and 106 is approximately 0.4987.