Answer:
The cutoff score is 71.
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.