Question

9) Assume that the random variable X is normally distributed, with mean = 90 and standard deviation o = 12. Compute the probability P(57 < X < 105).
A) 0.7888 B) 0.8944 C) 0.8914 D) 0.8819​

Answer 1

Answer:

Step-by-step explanation:

The mean, $\bar{x}$, is 90 and the standard deviation, $\sigma$, is 12.  We are looking for the probability that the variable X will fall between 57 and 105. We use the z-score table for this, AFTER we find the z scores. The formula to find the z-scores for us is:

$P(\frac{57-\bar{x}}{\sigma}\leq z\leq \frac{105-\bar{x}}{\sigma})$ and we fill in accordingly:

$P(\frac{57-90}{12}\leq z\leq \frac{105-90}{12})$ which simplifies to

$P(-2.75\leq z\leq 1.25)$ and we will break them up into 2 different sets as follows:

P(-2.75 ≤ z ≤ 0) + P(0 ≤ z ≤ 1.25)

and based on the fact that z scores are given from 0 on up, we are going to convert the first one by using the logic that if z is greater than -2.75 but less than 0, by symmetry, z is greater than 0 but less than 2.75:

P(0 ≤ z ≤ 2.75) + P(0 ≤ z ≤ 1.25) and we go to the z-score table.

Locate 2.7 down along the left side and move over til you're under the .05; that gives us the z-score for 2.75 which is .4970. Do the same for 1.25 to get a z-score of .3944. Add them together to get a final z-score that covers the range of values for X:

.4970 + .3944 = 0.8914