Question

The children in my subdivision were bored one summer and decided to hold a competition to see who could run to the end of the street in the shortest time. One hundred children participated, and the average finish time was 75 seconds. The SD was 20 seconds and the histogram of the times followed the normal curve. Estimate the percentage of children who finished the race in under 45 seconds.

Answer 1

Answer:

6.7% of children who finished the race in under 45 seconds.

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 75 seconds

Standard Deviation, σ =  20 seconds

We are given that the distribution of time of completion is a bell shaped distribution that is a normal distribution.

Formula:

$z_{score} = \displaystyle\frac{x-\mu}{\sigma}$

P(finished the race in under 45 seconds.)

P(x < 45)

$P( x < 45) = P( z < \displaystyle\frac{45 - 75}{20}) = P(z < -1.5)$

Calculation the value from standard normal z table, we have,  

$P(x < 45) = 0.067 = 6.7\%$

6.7% of children who finished the race in under 45 seconds.