Question

For a normal distribution with mean 47 and standard deviation 6, find the probability of obtaining a value less than 45 or greater than 49.

Answer 1

Answer:

0.7392 = 73.92% probability of obtaining a value less than 45 or greater than 49.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean 47 and standard deviation 6

This means that $\mu = 47, \sigma = 6$

Less than 45:

p-value of Z when X = 45, so:

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

$Z = \frac{45 - 47}{6}$

$Z = -0.3333$

$Z = -0.3333$ has a p-value of 0.3696.

More than 49:

1 subtracted by the p-value of Z when X = 49. So

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

$Z = \frac{49 - 47}{6}$

$Z = 0.3333$

$Z = 0.3333$ has a p-value of 0.6304.

1 - 0.6304 = 0.3996

Less than 45 or greater than 49:

2*0.3696 = 0.7392

0.7392 = 73.92% probability of obtaining a value less than 45 or greater than 49.