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%.

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

its 4 because 2 times 2 is 4, if you double the number two you get four

$9.89

step-by-step explanation:

9.90 = p x .06 x .25 (3/12 changed to a decimal)

9.90 = p x 0.015

9.90 - 0.015 = p

9.89 = p