Question

Assume that the random variable X is normally​ distributed, with mean mu equals 80μ=80 and standard deviation sigma equals 20.σ=20. Compute the probability ​P(Xgreater than>9696​).

Answer 1

`P(X \geq 86) = 1 - 0.7881 = 0.2119`

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 = (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 problem, we have that:

`\mu = 80, \sigma = 20`

Compute the probability ​P(Xgreater than>96​).

This is 1 subtracted by the pvalue of Z when X = 96. So

`Z = (X - \mu)/(\sigma)`

`Z = (96 - 80)/(20)`

`Z = 0.8`

`Z = 0.8` has a pvalue of 0.7881

`P(X \geq 86) = 1 - 0.7881 = 0.2119`