Question

Assume the random variable x is normally distributed with mean mu equals 50μ=50 and standard deviation sigma equals 7σ=7. find the indicated probability. upper p left parenthesis x greater than 34 right parenthesisp(x>34)

Answer 1

P(X > 34) = 1 - 0.011 = 0.989

Explanation:

Problems of normally distributed samples can be 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 = 50, \sigma = 7`

The probability that the problem asks is 1 subtracted by the pvalue of Z when X = 34. So:

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

`Z = (34 - 50)/(7)`

`Z = -2.29`

`Z = -2.29` has a pvalue of 0.011.

So

P(X > 34) = 1 - 0.011 = 0.989

Answer 2

To get the P(x>34) we proceed as follows:
z-score is given by:
z=(x-mu)/sig
z=(34-50)/7
z=-2.29
Thus:
P(x>34)=1-P(x<34)=1-0.0110=0.989
Answer: p(X>34)=0.989