Question

Assume that the random variable X is normally​ distributed, with mean muequals45 and standard deviation sigmaequals10. Compute the probability ​P(57 > than X less than or = 69​).

Answer 1

Answer:

0.1069 = 10.69%

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this question:

$\mu = 45, \sigma = 10$

Between 57 and 69

This is the pvalue of Z when X = 69 subtracted by the pvalue of Z when X = 57. So

X = 69

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{69 - 45}{10}$

$Z = 2.4$

$Z = 2.4$ has a pvalue of 0.9918

X = 57

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{57 - 45}{10}$

$Z = 1.2$

$Z = 1.2$ has a pvalue of 0.8849

0.9918 - 0.8849 = 0.1069 = 10.69%