In order to qualify for a police academy, applicants are given a test of physical fitness. The scores are normally distributed with a mean of 60 and a standard deviation of 12. If only the top 20% of the applicants are selected find the cutoff score. Round value calculations to 2 decimal places and final answer to the nearest whole number

The cutoff score is 71.

Step-by-step explanation:

We are given that the scores are normally distributed with a mean of 60 and a standard deviation of 12.

If only the top 20% of the applicants are selected we have to find the cut off score.

Let X = sample mean daily precipitation

So, X ~ Normal()

The z score probability distribution for normal distribution is given by;

                                 Z =  ~ N(0,1)

where, = population mean score = 60

            = standard deviation = 12

Now, we have to find the cut off score at which the top 20% of the applicants are selected, that is;

     P(X x) = 0.20           {where x is the required cut off score}

     P( ) = 0.20

     P(Z ) = 0.20

Now, in the z table the critical value of x that represents the top 20% of the area is given as 0.8416, i.e;

               = 0.8416

                 x = 60 + 10.09 = 70.09 ≈ 71

Hence, the cutoff score is 71.

use this formula:

p(a\b) = p(a and b)/p(b).

take it from here.