Question

Consider a bell-shaped symmetric distribution with mean of 16 and standard deviation of 1.5. Approximately what percentage of data lie between 13 and 19?

Answer 1

Answer: 95.45 %

Step-by-step explanation:

Given : The distribution is bell shaped , then the distribution must be normal distribution.

Mean : $\mu=\ 16$

Standard deviation :$\sigma= 1.5$

The formula to calculate the z-score :-

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

For x = 13

$z=\dfrac{13-16}{1.5}=-2$

For x = 19

$z=\dfrac{19-16}{1.5}=2$

The p-value = $P(-2$

$0.9772498-0.0227501=0.9544997\approx0.9545$

In percent, $0.9545\times100=95.45\%$

Hence, the percentage of data lie between 13 and 19 = 95.45 %