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Zielflug [23.3K]
3 years ago
15

Assume that the random variable X is normally distributed, with mean p = 100 and standard deviation o = 15. Compute the

Mathematics
1 answer:
mamaluj [8]3 years ago
7 0

Answer:

P(X > 112) = 0.21186.

Step-by-step explanation:

We are given that the random variable X is normally distributed, with mean \mu = 100 and standard deviation \sigma = 15.

Let X = <u><em>a random variable</em></u>

The z-score probability distribution for the normal distribution is given by;

                               Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean = 100

            \sigma = standard deviaton = 15

Now, the probability that the random variable X is greater than 112 is given by = P(X > 112)

      P(X > 112) = P( \frac{X-\mu}{\sigma} > \frac{112-100}{15} ) = P(Z > 0.80) = 1- P(Z \leq 0.80)

                                                       = 1 - 0.78814 = <u>0.21186</u>  

The above probability is calculated by looking at the value of x = 0.80 in the z table which has an area of 0.78814.

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