Question

a physical fitness association is including the mile run in its high school fitness test. the times for high school boys in this event are known to be normally distributed. it is also known that the approximate middle 95% of boys will finish the run with times between 5 minutes 35 seconds and 10 minutes 5 seconds.

1. decide from this the mean mile run times for high school boys?

2. decide from this the approximate standard deviation of mile run times for high school boys?

3. if 347 boys at a particular high school run the mile, approximately how many of the boys do you expect to take longer than 6 min 5 sec?

Answer 1

Answer)

According to the empirical rule

If the data is normally distributed

Then 68% lies in between mean - s.d and mean + s.d

95% lies in between mean - 2*s.d and mean + 2*s.d

99.7% lies in between mean - 3*s.d and mean + 3*s.d

So, as 95% lies in between 5 minutes 35 seconds = 335 seconds and 10 minutes and 5 seconds (605 seconds)

So, 335 = mean - 2*s.d .... Equation 1

605 = mean + (2*s.d) ..... Equation 2

Adding both the equations we get

2*mean = (335+605)

Mean = 470

2)

Substituting 470 in equation 1

S.d = 67.5

3)

6 minutes 5 seconds = 365 seconds

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 470

S.d = 67.5

We need to find, P(x>365 )

Z = (365-470)/67.5 = -1.56

From z table, P(z>-1.56) = 0.9406

Expected value is given by n*p

= 347*0.9406

= 326.3882

~ 326