Answer :
Sure! Let's go through each part of the question step-by-step.
### a.) Find the mean and standard deviation of the total time for the team to complete their [tex]\(4 \times 100\)[/tex] meter relay.
Step 1: Finding the Mean
- We are given the mean times for each swimmer:
- Swimmer A: 58.4 seconds
- Swimmer B: 59.1 seconds
- Swimmer C: 57.8 seconds
- Swimmer D: 58.6 seconds
To find the total mean time for the relay, we add up the mean times of all the swimmers:
[tex]\[
\text{Total Mean} = 58.4 + 59.1 + 57.8 + 58.6 = 233.9 \text{ seconds}
\][/tex]
Step 2: Finding the Standard Deviation
- The standard deviations for each swimmer are given as:
- Swimmer A: 1.2 seconds
- Swimmer B: 1.1 seconds
- Swimmer C: 0.8 seconds
- Swimmer D: 0.9 seconds
Since the times are independent, we can find the total standard deviation by combining the variances (square of standard deviations) and taking the square root:
[tex]\[
\text{Variance (Total)} = (1.2)^2 + (1.1)^2 + (0.8)^2 + (0.9)^2 = 1.44 + 1.21 + 0.64 + 0.81 = 4.1
\][/tex]
[tex]\[
\text{Total Standard Deviation} = \sqrt{4.1} \approx 2.02 \text{ seconds}
\][/tex]
### b.) Probability the team will beat their best time of 231.8 seconds
- The mean total time is 233.9 seconds with a standard deviation of approximately 2.02 seconds.
- To find the probability of the team beating 231.8 seconds, calculate the z-score:
[tex]\[
z = \frac{231.8 - 233.9}{2.02} \approx -1.04
\][/tex]
Using the properties of the normal distribution, find the probability:
- A z-score of [tex]\(-1.04\)[/tex] corresponds to 0.1498 in the standard normal distribution table.
- Therefore, the probability of the team finishing faster than 231.8 seconds is:
[tex]\[
1 - 0.1498 = 0.8502
\][/tex]
### c.) Probability of Swimmer D completing faster than Swimmer A
- Swimmer A's mean time is 58.4 seconds with a standard deviation of 1.2 seconds.
- Swimmer D's mean time is 58.6 seconds with a standard deviation of 0.9 seconds.
Calculate the z-score for the difference:
- The difference in their mean times:
[tex]\[
\text{Mean Difference} = 58.4 - 58.6 = -0.2
\][/tex]
- The standard deviation of the difference, assuming independence, is calculated as follows:
[tex]\[
\text{Std. Dev (Difference)} = \sqrt{(1.2)^2 + (0.9)^2} = \sqrt{1.44 + 0.81} = \sqrt{2.25} = 1.5
\][/tex]
- Calculate the z-score for the difference:
[tex]\[
z = \frac{-0.2}{1.5} \approx -0.13
\][/tex]
Using the normal distribution table, the probability of obtaining a z-score less than [tex]\(-0.13\)[/tex] is approximately 0.4469, meaning Swimmer D has about a 44.7% chance of completing his lap faster than Swimmer A.
I hope this helps you understand how to approach and solve each part of the question!
### a.) Find the mean and standard deviation of the total time for the team to complete their [tex]\(4 \times 100\)[/tex] meter relay.
Step 1: Finding the Mean
- We are given the mean times for each swimmer:
- Swimmer A: 58.4 seconds
- Swimmer B: 59.1 seconds
- Swimmer C: 57.8 seconds
- Swimmer D: 58.6 seconds
To find the total mean time for the relay, we add up the mean times of all the swimmers:
[tex]\[
\text{Total Mean} = 58.4 + 59.1 + 57.8 + 58.6 = 233.9 \text{ seconds}
\][/tex]
Step 2: Finding the Standard Deviation
- The standard deviations for each swimmer are given as:
- Swimmer A: 1.2 seconds
- Swimmer B: 1.1 seconds
- Swimmer C: 0.8 seconds
- Swimmer D: 0.9 seconds
Since the times are independent, we can find the total standard deviation by combining the variances (square of standard deviations) and taking the square root:
[tex]\[
\text{Variance (Total)} = (1.2)^2 + (1.1)^2 + (0.8)^2 + (0.9)^2 = 1.44 + 1.21 + 0.64 + 0.81 = 4.1
\][/tex]
[tex]\[
\text{Total Standard Deviation} = \sqrt{4.1} \approx 2.02 \text{ seconds}
\][/tex]
### b.) Probability the team will beat their best time of 231.8 seconds
- The mean total time is 233.9 seconds with a standard deviation of approximately 2.02 seconds.
- To find the probability of the team beating 231.8 seconds, calculate the z-score:
[tex]\[
z = \frac{231.8 - 233.9}{2.02} \approx -1.04
\][/tex]
Using the properties of the normal distribution, find the probability:
- A z-score of [tex]\(-1.04\)[/tex] corresponds to 0.1498 in the standard normal distribution table.
- Therefore, the probability of the team finishing faster than 231.8 seconds is:
[tex]\[
1 - 0.1498 = 0.8502
\][/tex]
### c.) Probability of Swimmer D completing faster than Swimmer A
- Swimmer A's mean time is 58.4 seconds with a standard deviation of 1.2 seconds.
- Swimmer D's mean time is 58.6 seconds with a standard deviation of 0.9 seconds.
Calculate the z-score for the difference:
- The difference in their mean times:
[tex]\[
\text{Mean Difference} = 58.4 - 58.6 = -0.2
\][/tex]
- The standard deviation of the difference, assuming independence, is calculated as follows:
[tex]\[
\text{Std. Dev (Difference)} = \sqrt{(1.2)^2 + (0.9)^2} = \sqrt{1.44 + 0.81} = \sqrt{2.25} = 1.5
\][/tex]
- Calculate the z-score for the difference:
[tex]\[
z = \frac{-0.2}{1.5} \approx -0.13
\][/tex]
Using the normal distribution table, the probability of obtaining a z-score less than [tex]\(-0.13\)[/tex] is approximately 0.4469, meaning Swimmer D has about a 44.7% chance of completing his lap faster than Swimmer A.
I hope this helps you understand how to approach and solve each part of the question!
