Mathematics: 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).

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Assume that the random variable X is normally distributed,

Ответ:

yejinschoi2463

20.01.2022

Mathematics

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

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 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

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

$Z = \frac{96 - 80}{20}$

Z = 0.8

Z = 0.8 has a pvalue of 0.7881

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

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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).

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