Question

After the physical training required during World War II, the distribution of mile run times for male students at the University of Illinois was approximately Normal with mean 7.11 minutes and standard deviation 0.74 minutes. What proportion of these students could run a mile in 7 minutes or less?

Answer 1

To determine the proportion of male students at the University of Illinois who could run a mile in 7 minutes or less, we calculate the z-score and use the standard normal distribution table.

Step-by-step explanation:

To find the proportion of male students at the University of Illinois who could run a mile in 7 minutes or less, we need to calculate the z-score and use the standard normal distribution. The formula for z-score is:

z = (x - mean) / standard deviation

Substituting the given values, we get:

z = (7 - 7.11) / 0.74 = -0.14

Using the standard normal distribution table, we can find the proportion corresponding to a z-score of -0.14, which is approximately 0.4442. Therefore, approximately 44.42% of the male students could run a mile in 7 minutes or less.

Answer 2

`(1101)/(2500)`

Step-by-step explanation:

Given,

Mean,
`\mu=7.11` minutes,

Standard deviation =
`\sigma = 0.74` minutes,

Let x represents the run time,

Thus, the probability of students who could run a mile in 7 minutes or less

= P( x ≤ 7 )

`=P((x-\mu)/(\sigma) \leq (x-\mu)/(\sigma))`

`=P(z\leq (7-7.11)/(0.74))`

`=P(z\leq -0.15 )`

By the normal distribution table,

`= 0.4404`

`=(4404)/(10000)`

`=(1101)/(2500)`

Hence, the proportion of students who could run a mile in 7 minutes or less is about
`(1101)/(2500)`