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?

Respuesta :

Answer: 95.45 %

Step-by-step explanation:

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

Mean : [tex]\mu=\ 16[/tex]

Standard deviation :[tex]\sigma= 1.5[/tex]

The formula to calculate the z-score :-

[tex]z=\dfrac{x-\mu}{\sigma}[/tex]

For x = 13

[tex]z=\dfrac{13-16}{1.5}=-2[/tex]

For x = 19

[tex]z=\dfrac{19-16}{1.5}=2[/tex]

The p-value = [tex]P(-2<z<2)=P(z<2)-P(z<-2)[/tex]

[tex]0.9772498-0.0227501=0.9544997\approx0.9545[/tex]

In percent, [tex]0.9545\times100=95.45\%[/tex]

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