Answer: the qualifying time in seconds is about 25.3
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