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amid [387]
3 years ago
7

Assume that the random variable X is normally distributed, with mean 60 and standard deviation 16. Compute the probability P(X &

lt; 80). Group of answer choices
Mathematics
1 answer:
elixir [45]3 years ago
5 0

Answer:

P(X < 80) = 0.89435.

Step-by-step explanation:

When the distribution is normal, we use 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, we have that:

\mu = 60, \sigma = 16

P(X < 80)

This is the pvalue of Z when X = 80. So

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

Z = \frac{80 - 60}{16}

Z = 1.25

Z = 1.25 has a pvalue of 0.89435.

So

P(X < 80) = 0.89435.

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