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

a.   since the profit made in 2012 is equal to the profit made in 2014, the profit is symmetric about the line y = 120,000.

step-by-step explanation:

fist on the left goes with the third on the right. second on the left goes with the fourth on the right. third on the left goes with the second on the right. the fourth on the left goes with first on the right