Ask Question
11 October, 23:06

Scores on an english test are normally distributed with a mean of 31.5 and a standard deviation of 7.3. find the score that separates the top 59% from the bottom 41%.

+3
Answers (1)
  1. 12 October, 02:30
    0
    Let X be the score on an english test which is normally distributed with mean of 31.5 and standard deviation of 7.3

    μ = 31.5 and σ = 7.3

    Here we have to find score that separates the top 59% from the bottom 41%

    So basically we have to find here x value such that area above it is 59% and below it is 49%

    This is same as finding z score such that probability below z score is 0.49 and above probability is 0.59

    P (Z < z) = 0.49

    Using excel function to find the z score for probability 0.49 we get

    z = NORM. S. INV (0.49)

    z = - 0.025

    It means for z score - 0.025 area below it is 41% and above it is 59%

    Now we will convert this z score into x value using given mean and standard deviation

    x = (z * standard deviation) + mean

    x = (-0.025 * 7.3) + 31.5

    x = 31.6825 ~ 31.68

    The score that separates the top 59% from the bottom 41% is 31.68
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Scores on an english test are normally distributed with a mean of 31.5 and a standard deviation of 7.3. find the score that separates the ...” in 📙 Mathematics if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers