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.
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?
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.
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
To answer these questions, we need to use the properties of a normal distribution and the given mean and standard deviation.
a.) To find the number of swimmers who will finish in under 49 seconds, we need to calculate the area under the normal distribution curve to the left of 49 seconds.
First, we need to standardize the value 49 by calculating the z-score. The z-score formula is:
z = (x - mean) / standard deviation
z = (49 - 55) / 5
z = -6 / 5
z = -1.2
Now, we can use a standard normal distribution table or calculator to find the area or probability associated with this z-score. The standard normal distribution table provides the area to the left of the z-score.
Approximately, the area to the left of -1.2 in the standard normal distribution is 0.1151 or 11.51%.
To find the number of swimmers who finish in under 49 seconds, we need to multiply this probability by the total number of swimmers in the population.
Number of certificates needed = Probability * Total number of swimmers
= 0.1151 * 50 (since there are 50 swimmers)
= 5.755 ≈ 6 certificates
So, approximately 6 certificates will be needed for swimmers who finish in under 49 seconds.
b.) To find the finishing time for a swimmer to be in the fastest 2% of the distribution, we need to calculate the z-score corresponding to this percentile.
Since the fastest 2% corresponds to the area to the right of this z-score, we need to find the z-score that corresponds to an area of 0.98 (1 - 0.02) in the standard normal distribution.
Using a standard normal distribution table or calculator, we can find that the z-score corresponding to an area of 0.98 is approximately 2.05.
Now, we can use the z-score formula to find the finishing time:
z = (x - mean) / standard deviation
2.05 = (x - 55) / 5
Solving for x:
2.05 * 5 = x - 55
10.25 = x - 55
x = 10.25 + 55
x = 65.25
Therefore, a swimmer must finish in approximately 65.25 seconds to be in the fastest 2% of the distribution of finishing times.