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

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

Answer 2

Answer: 22.95 %

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=(x-\mu)/(\sigma)`

For x=360

`z=(360-350)/(25)=0.4`

For x=380

`z=(380-350)/(25)=1.2`

The p-value =
`P(360<X<380)`

`=P(0.4<z<1.2)=P(1.2)-P(0.4)\\\\= 0.8849303-0.6554217=0.2295086\approx0.2295`

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

`0.2295*100=22.95\%`