Question

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

Answer 1

Answer:

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($\mu=60,\sigma^{2} =12^{2}$)

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

                                 Z = $\frac{ X-\mu}{\sigma } }$  ~ N(0,1)

where, $\mu$ = population mean score = 60

           $\sigma$ = 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 $\geq$ x) = 0.20           {where x is the required cut off score}

     P( $\frac{ X-\mu}{\sigma } }$ $\geq$ $\frac{x-60}{12 } }$ ) = 0.20

     P(Z $\geq$ $\frac{x-60}{12 } }$ ) = 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;

             $\frac{x-60}{12 } }$  = 0.8416

              ${x-60}}= 0.8416 \times 12$

                 x = 60 + 10.09 = 70.09 ≈ 71

Hence, the cutoff score is 71.