The overhead reach distances of adult females are normally distributed with a mean of 200 cm200 cm and a standard deviation of 8 cm8 cm. a. find the probability that an individual distance is greater than 209.30209.30 cm. b. find the probability that the mean for 1515 randomly selected distances is greater than 198.20 cm.198.20 cm. c. why can the normal distribution be used in part (b), even though the sample size does not exceed 30?

A) 0.123; B) 0.8078; C) Because the population is normally distributed.

Step-by-step explanation:

For part A,

We first calculate the z-score, which tells us how many standard deviations from the mean our score is.

Since we are finding the z-score of an individual score and not a sample, we use the formula

Our score, X, is 209.3; our mean, μ, is 200; and our standard deviation, σ, is σ.  This gives us

We look this value up in a z-table.  However, we must round it to the nearest hundredth first; this is 1.16.  From a z-table, we get that the area under the curve to the left of, or less than or equal to, this score is 0.8770.

However we want the probability that the value is greater than this; this means we subtract our found probability from 1:

1-0.8770 = 0.123

For part B,

Again we first calculate the z-score.  However this time we are finding the probability of a mean of a sample rather than an individual score; this means we use the formula

Our X is 198.20; our μ is still 200, and our σ is still 8; our value of n is our sample size, 15:

Looking up this value in a z-table, we get 0.1922.

However, we want the probability that the area, or probability, is greater than this; so we subtract from 1:

1-0.1922 = 0.8078

For part C,

When a population is normally distributed, this means that a sample taken from this population will also be normal; this means we can use the normal distribution.

3/6   and 9/18

step-by-step explanation:

your answer would be 12