Consider a test that has a normal distribution, a mean of 100, and a standard deviation of 14. how high of a score would a person need to be in the top 1%

A test has normal distribution with mean μ =100 and standard deviation σ =14

Let X be the score obtained by person in test. Here we have to find test score x such that person would be in top 1%

It means x is such that probability above x is 0.01 and below is 0.9

So we will find z score such that area below z is 0.9

In z score table there is no accurate probability value 0.9 so we take probability value close to 0.9 which is 0.8997 and corresponding z score is 1.28

Now we will find x value from z=1.28

x = (z * standard deviation) + mean

x = (1.28 * 14) + 100

x = 117.92 ~ 118

The score value for a person to be in top 1% is 118.

the answer is 48m^2!

step-by-step explanation: