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.) b. In what amount of time must a swimmer finish to be in the fastest 2% of the distribution of finishing times?
Repost with data
a) To determine the approximate number of certificates needed for swimmers who finish in under 49 seconds, we need to calculate the proportion of the population that falls under this threshold.
Step 1: Calculate the z-score for the value of 49 seconds using the formula:
Z = (X - μ) / σ
where X is the value we are interested in (49 seconds), μ is the mean (55 seconds), and σ is the standard deviation (5 seconds).
Z = (49 - 55) / 5
Z = -6 / 5
Z = -1.2
Step 2: Determine the proportion of the population that falls below the z-score of -1.2 using a standard normal distribution table or a calculator.
From the standard normal distribution table, the area to the left of -1.2 is approximately 0.1151.
Step 3: Calculate the approximate number of swimmers who will finish in under 49 seconds by multiplying the proportion from step 2 by the total number of swimmers (50).
Number of certificates needed = 0.1151 * 50
Number of certificates needed ≈ 5.76
Since you cannot have a fraction of a certificate, you would round up the number to the nearest whole number. Therefore, approximately 6 certificates will be needed.
b) To determine the finishing time that corresponds to the fastest 2% of the distribution, we need to find the z-score that corresponds to this percentile.
Step 1: Find the z-score that corresponds to the 98th percentile by subtracting the area of 2% from 1 (since we want the area to the left of the z-score).
This can also be referred to as finding the z-score for a cumulative probability of 0.98.
Step 2: Use a standard normal distribution table, calculator, or software to find the z-score that corresponds to a cumulative probability of 0.98.
From the standard normal distribution table, the z-score corresponding to a cumulative probability of 0.98 is approximately 2.05.
Step 3: Convert the z-score back to the original value using the formula:
X = μ + (Z * σ)
where X is the value we are interested in (the finishing time), μ is the mean (55 seconds), σ is the standard deviation (5 seconds), and Z is the z-score (2.05).
Finishing time = 55 + (2.05 * 5)
Finishing time = 55 + 10.25
Finishing time ≈ 65.25 seconds
Therefore, a swimmer must finish in approximately 65.25 seconds to be in the fastest 2% of the distribution of finishing times.