Question

Consider a normal distribution with mean 23 and standard deviation 7. What is the probability a value selected at random from this distribution is greater than 23? (Round your answer to two decimal places.)

Answer 1

Answer:

0.5 = 50% probability a value selected at random from this distribution is greater than 23

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:

$\mu = 23, \sigma = 7$

What is the probability a value selected at random from this distribution is greater than 23?

This is 1 subtracted by the pvalue of Z when X = 23. So

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

$Z = \frac{23 - 23}{7}$

$Z = 0$

$Z = 0$ has a pvalue of 0.5

0.5 = 50% probability a value selected at random from this distribution is greater than 23

Answer 2

Answer:

P(X > 23) = 0.50 .

Step-by-step explanation:

We are given a normal distribution with mean 23 and standard deviation 7.

Let X = randomly selected value

So, X ~ N($\mu = 23,\sigma^{2} =7^{2}$)

The z score probability distribution is given by;

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

So, probability that a value selected at random from this distribution is greater than 23 = P(X > 23)

P(X > 23) = P( $\frac{X-\mu}{\sigma}$ > $\frac{23-23}{7}$ ) = P(Z > 0) = 0.50 {using z table)

Therefore, probability that a value selected at random from this distribution is greater than 23 is 50% .