To qualify for security officers’ training recruits are tested for stress tolerance. The scores are normally distributed, with a mean of 62 and a standard deviation of 8.If only the top 15% of recruits are selected, find the cutoff score.

72.48

Step-by-step explanation:

Knowing that the z value has the following equation:

z = (x - m) / d

Let x be the cut-off point, m the arithmetic mean and d the standard deviation, we want to solve for the cut-off point so it would look like this:

x = d * z + m

d = 8

m = 62

We calculate z using the attached normal distribution table, we look for when z is 0.85, and we can see that in 0.8508 the value is 1.04.

So we have to:

x = 62 * 1.04 + 8 = 72.48

In other words, the cut point is approximately 72.5

or

step-by-step explanation: