1. The finishing times for the population of all swimmers performing the 100-meter butterfly are normally distributed with a mean ƒÝ of 55 seconds and a standard deviation ƒã of 5 seconds.

Question

a.) The sponsors decide to give certificates to all swimmers who finish in under 49 seconds. If there are 50 swimmers in a 100-meter butterfly, approximately how many certificates will be needed?

b.) In what amount of time must a swimmer finish to be in the fastest 2% of the distribution of finishing times?

Answer 1

Use z-scores and the normal distribution table for these questions.

Formula:
z = (x - mean)/sd
... sd = standard deviation

Using your data:
z = (49 - 55)/5 = ?
Finish the calculation for the z-score (hint: the score will be negative).

Use a normal distribution table to determine the proportion who finish under 49 seconds. Multiply that proportion by 50 to determine how many certificates will be needed.

For part b), convert the fastest 2% into a z-score, then use the z-score formula to solve for x (which will represent the time).

I hope this will give you a few hints to help get you started.

Answer 2

mean = 55
sigma = 5

(55 - 49)/5 = 1.2 sigma below mean time
from normal distribution table about .115 or 11.5% are below that z value
.115*50 = 5.75, buy 6 certificates

.02 is at z = 2 in table
so (55-t)/5 = 2
55 -t = 10
t =45 seconds