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Chapter 6: Problem 63

Swim team Hanover High School has the best women鈥檚 swimming team in the region. The 400- meter freestyle relay team is undefeated this year. In the 400-meter freestyle relay, each swimmer swims 100 meters. The times, in seconds, for the four swimmers this season are approximately Normally distributed with means and standard deviations as shown. Assuming that the swimmer鈥檚 individual times are independent, find the probability that the total team time in the 400-meter freestyle relay is less than 220 seconds. Follow the four-step process. Swimmer Mean Std. dev. Wendy 55.2 2.8 Jill 58.0 3.0 Carmen 56.3 2.6 Latrice 54.7 2.7

Short Answer

Expert verified
The probability is approximately 23.7%.

Step by step solution

01

Define the Random Variables

For each swimmer, define a random variable.Let \( X_1 \) be the time Wendy takes, \( X_2 \) be the time Jill takes, \( X_3 \) be the time Carmen takes, and \( X_4 \) be the time Latrice takes.\( X_1 \sim N(55.2, 2.8^2), \; X_2 \sim N(58.0, 3.0^2), \; X_3 \sim N(56.3, 2.6^2), \; X_4 \sim N(54.7, 2.7^2) \).
02

Find the Distribution of Total Time

The total time \( T \) is the sum of the times for the four swimmers: \[ T = X_1 + X_2 + X_3 + X_4 \].The mean of \( T \) is the sum of their individual means: \( \mu_T = 55.2 + 58.0 + 56.3 + 54.7 = 224.2 \).The variances add because the times are independent: \( \sigma_T^2 = 2.8^2 + 3.0^2 + 2.6^2 + 2.7^2 = 34.34 \).Thus, \( T \sim N(224.2, \sqrt{34.34}) \).
03

Find the Probability

Find \( P(T < 220) \) using the standard normal distribution.Convert to a standard normal variable \( Z \): \( Z = \frac{T - 224.2}{\sqrt{34.34}} \approx \frac{220 - 224.2}{5.86} \approx -0.716 \).Use a standard normal distribution table or calculator to find \( P(Z < -0.716) \).
04

Interpret the Result

From the standard normal table, \( P(Z < -0.716) \approx 0.237 \).This means there is a 23.7% chance that the team finishes the relay in less than 220 seconds.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Distribution
The normal distribution is a key concept in statistics, especially when dealing with probabilities concerning continuous random variables. It is often portrayed as a bell-shaped curve, symmetric around its mean. Many natural phenomena follow this distribution, which makes it highly significant in statistical analysis.

In the context of the swim team problem, each swimmer鈥檚 time is considered to be normally distributed. This means that 68% of their swimming times fall within one standard deviation from the mean. About 95% fall within two standard deviations, and 99.7% within three standard deviations.
  • For Wendy, the normal distribution is centered around 55.2 seconds with a spread of 2.8 seconds.
  • Jill鈥檚 distribution centers at 58.0 seconds with a spread of 3.0 seconds.
  • Carmen has a center of 56.3 seconds and a spread of 2.6 seconds.
  • Latrice centers at 54.7 seconds with a spread of 2.7 seconds.
These individual distributions help calculate the probability of the swim team's total time being less than 220 seconds.
Random Variables
Random variables are fundamental components in probability and statistics that represent possible outcomes of a random phenomenon. Each swimmer's time in the relay is treated as a random variable. They are unpredictable to some extent, but they follow a statistical pattern.

In this exercise:
  • Wendy鈥檚 swim time is denoted as the random variable \( X_1 \).
  • Jill's time is represented as \( X_2 \).
  • Carmen's time is noted as \( X_3 \).
  • Latrice's time is denoted as \( X_4 \).
Each of these random variables is described by a normal distribution with specific means and variances. They are incredibly helpful in modeling and predicting the swim times by using probability distributions.
Independence in Statistics
Independence is an important concept in statistics, indicating that one event or variable does not influence or provide any information about another. In the context of this exercise, the independence of each swimmer's time means that knowing one swimmer鈥檚 time does not help predict another鈥檚.

This assumption simplifies calculations significantly because the variances of the random variables can be added directly, without any additional calculation for covariances. In our swim team scenario, the times of Wendy, Jill, Carmen, and Latrice are considered independent, which allows the straightforward summing of variances to find the total distribution of the relay time.

This independence assumption is essential for computing the combined effect of individual performances on the relay, such as finding the probability of the total time being under 220 seconds.
Standard Deviation
Standard deviation is a measure of how spread out numbers are from their average value. It's one of the most common ways to quantify the amount of variation or dispersion in a set of data.

For the swimming times:
  • Wendy鈥檚 standard deviation is 2.8 seconds.
  • Jill has a standard deviation of 3.0 seconds.
  • Carmen鈥檚 is 2.6 seconds.
  • Latrice鈥檚 is 2.7 seconds.
These values indicate how much the swimmers鈥� times vary around their means. A smaller standard deviation means that the swimmer's times are consistently close to the mean, while a larger standard deviation indicates greater variability.

In the context of the relay, understanding the standard deviation helps us to grasp how much overall variability we can expect in the total time, which directly impacts the probability of achieving a certain relay time, like for under 220 seconds.

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Most popular questions from this chapter

A sample survey contacted an SRS of 663 registered voters in Oregon shortly after an clection and asked respondents whether they had voted. Voter records show that 56\(\%\) of registered voters had actually voted. We will see later that in repeated random samples of size 663 , the proportion in the sample who voted (call this proportion With vary according to the Normal distribution with mean \(\mu=0.56\) and standard deviation \(\sigma=0.019\) (a) If the respondents answer truthfully, what is \(\mathrm{P}(0.52 \leq V \leq 0.60)\) ? This is the probability that the sample proportion \(V\) estimates the population proportion 0.56 within \(\pm 0.04\) (b) In fact, 72\(\%\) of the respondents said they had voted \((V=0.72) .\) If respondents answer truthfully, what is \(P(V \geq 0.72) ?\) This probability is so small that it is good evidence that some people who did not vote claimed that they did vote.

A deck of cards contains 52 cards, of which 4 are aces. You are offered the following wager: Draw one card at random from the deck. You win \(\$ 10\) if the card drawn is an ace. Otherwise, you lose \(\$ 1 .\) If you make this wager very many times, what will be the mean amount you win? (a) About \(-\$ 1\) , because you will lose most of the time. (b) About \(\$ 9,\) because you win \(\$ 10\) but lose only \(\$ 1\) . (c) About \(-\$ 0.15 ;\) that is, on average you lose about 15 cents. (d) About \(\$ 0.77 ;\) that is, on average you win about 77 cents. (e) About \(\$ 0,\) because the random draw gives you a fair bet.

ITBS scores The Normal distribution with mean \(\mu=6.8\) and standard deviation \(\sigma=1.6\) is a good description of the lowa Test of Basic Skills (ITBS) vocabulary scores of seventh-grade students in Gary, Indiana. Call the score of a randomly chosen student \(X\) for short. Find \(P(X \geq 9)\) and interpret the result. Follow the four-step process.

Random numbers Let Y be a number between 0 and 1 produced by a random number generator. Assuming that the random variable Y has a uniform distribution, find the following probabilities: (a) \(P(Y \leq 0.4)\) (b) \(P(Y < 0.4)\) (c) \(P(0.1 < Y \leq 0.15 \text { or } 0.77 \leq Y < 0.88)\)

Exercises 27 and 28 refer to the following setting. Choose an American household at random and let the random variable X be the number of cars (including SUVs and light trucks) they own. Here is the probability model if we ignore the few households that own more than 5 cars: Number of cars X: 012345 Probability: 0.09 0.36 0.35 0.13 0.05 0.02 A housing company builds houses with two-car garages. What percent of households have more cars than the garage can hold? (a) 13\(\%\) (b) 20\(\%\) (c) 45\(\%\) (d) 55\(\%\) (e) 80\(\%\)
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