Personal best finishing times for a particular race in high school track meets are normally distributed with mean 24.6 seconds and standard deviation .64 seconds. If the qualifying time for this event for the regional championship is set so that the top 15 percent of all runners qualify, then that qualifying time in seconds is about:

the qualifying time in seconds is about 25.3

Step-by-step explanation:

Since the personal best finishing times for a particular race in high school track meets are normally distributed, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = personal best finishing times for a particular race.

µ = mean finishing time

σ = standard deviation

From the information given,

µ = 24.6 seconds

σ = 0.64 seconds

The probability value for the top 15% of finishing time for runners to qualify would be (1 - 15/100) = (1 - 0.15) = 0.85

Looking at the normal distribution table, the z score corresponding to the probability value is 1.04

Therefore,

1.04 = (x - 24.6)/0.64

Cross multiplying by 0.64, it becomes

1.04 × 0.64 = x - 24.6

0.6656 = x - 24.6

x = 0.6656 + 24.6

x = 25.3 seconds

d no solution

step-by-step explanation: