Question

Suppose a variable has a normal distribution with mean 67 and standard deviation 4. What percentage of the distribution is less than 75? (Use z-score.)

Answer 1

Answer:

The percentage of the distribution is less than 75 is 97.72%.

Step-by-step explanation:

Given,

Mean of the distribution,

$\mu=67$

Standard deviation,

$\sigma = 4$

Thus, the z-score of the score 75,

$z=\frac{x-\mu}{\sigma}$

$=\frac{75-67}{4}$

$=\frac{8}{4}$

$=2$

With the help of z-score table,

$P(x$

Hence, the percentage of the distribution is less than 75 is 97.72%.