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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?

User Crawford
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2 Answers

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

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\%

User Rao Sahab
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7.4k points