The IQ scores are normally distributed with a mean of 100 and a SD of 15.

A) What is the probability that a randomly person has an IQ b/t 85-115? Show work
B) Find the 90th percentile of the IQ distribution?
C) If a random sample of 100 people are selected, what is the SD of the sample mean?

You have to find the z-score.

z =(85-100)/15

z=(115 - 100)/15

Then you have to use a z-table or a calculator to find the area between the two z scores

You have to look at the table and find the z-score for the 90th percentile.

once you have the zscore

z = (x-100)/15

A) To find the probability that a randomly selected person has an IQ between 85 and 115, we need to calculate the z-scores and then use the z-table.

First, we calculate the z-score for a person with an IQ of 85:
z = (85 - 100) / 15 = -1

Next, we calculate the z-score for a person with an IQ of 115:
z = (115 - 100) / 15 = 1

Using a standard normal distribution table, we can find the cumulative probability for these z-scores.

The probability of a z-score of -1 or lower is 0.1587, and the probability of a z-score of 1 or lower is 0.8413.

To find the probability between these z-scores, we subtract the lower probability from the higher probability:
P(85 < X < 115) = 0.8413 - 0.1587 = 0.6826

Therefore, the probability that a randomly selected person has an IQ between 85 and 115 is 0.6826.

B) To find the 90th percentile of the IQ distribution, we need to find the z-score for which 90% of the data falls below it.

Using a standard normal distribution table, we find the z-score that corresponds to a cumulative probability of 0.90, which is 1.28.

Now, we can find the IQ value corresponding to this z-score:
IQ = (z * SD) + mean
= (1.28 * 15) + 100
= 19.2 + 100
= 119.2

Therefore, the 90th percentile of the IQ distribution is an IQ score of 119.2.

C) The standard deviation of the sample mean (also known as the standard error) can be calculated using the formula:

SE = SD / sqrt(n)

where SD is the population standard deviation (15 in this case) and n is the sample size (100 in this case).

SE = 15 / sqrt(100)
= 15 / 10
= 1.5

Therefore, the standard deviation of the sample mean for a sample size of 100 is 1.5.

A) To find the probability that a randomly selected person has an IQ between 85 and 115, we need to calculate the area under the normal distribution curve within that range.

Step 1: Calculate the z-scores for the lower and upper limits of the range.
For the lower limit (85):
z1 = (85 - 100) / 15 = -1

For the upper limit (115):
z2 = (115 - 100) / 15 = 1

Step 2: Look up the corresponding cumulative probabilities for the z-scores using a standard normal distribution table or a calculator. The cumulative probability for z1 is 0.1587, and for z2, it is 0.8413.

Step 3: Calculate the probability by finding the difference between the cumulative probabilities:
P(85 ≤ X ≤ 115) = P(X ≤ 115) - P(X ≤ 85)
= 0.8413 - 0.1587
= 0.6826

Therefore, the probability that a randomly selected person has an IQ between 85 and 115 is 0.6826, or 68.26%.

B) The 90th percentile refers to the value below which 90% of the data falls. To find it, we need to calculate the z-score that corresponds to the 90th percentile.

Step 1: Find the z-score corresponding to a cumulative probability of 0.90. You can use a standard normal distribution table or a calculator to determine this. The z-score corresponding to a cumulative probability of 0.90 is approximately 1.28.

Step 2: Calculate the IQ value using the z-score formula:
X = (Z * SD) + mean
= (1.28 * 15) + 100
= 19.2 + 100
= 119.2

Therefore, the 90th percentile of the IQ distribution is 119.2.

C) When calculating the standard deviation (SD) of the sample mean, we use the formula: SD(sample mean) = SD / sqrt(n), where n is the sample size.

Given that the SD of the IQ scores population is 15, and the sample size is 100, we can compute the SD of the sample mean:

SD(sample mean) = 15 / sqrt(100)
= 15 / 10
= 1.5

Therefore, the SD of the sample mean is 1.5.