Question

Scores on a standardized test are normally distributed with a mean of 350 and a standard deviation of 25. Approximately what percent of students scored between 360 and 380?

Answer 1

Answer: 22.95 %

Step-by-step explanation:

Given : Scores on a standardized test are normally distributed with

Mean : $\mu=350$

Standard deviation : $\sigma =25$

The formula to find z-score:-

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

For x=360

$z=\dfrac{360-350}{25}=0.4$

For x=380

$z=\dfrac{380-350}{25}=1.2$

The p-value = $P(360$

$=P(0.4$

In percent , the percent of students scored between 360 and 380 =

$0.2295\times100=22.95\%$

Answer 2

(360) = 10/25 = .4
z(380) = 30/25 = 1.2
P(350 < x < 380) = normalcdf(.4, 1.2)