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Sagot :
To solve this problem, we need to understand the probability distribution properties and use the Z-score, which is a measure of how many standard deviations an element is from the mean. Here’s a detailed step-by-step solution:
### Problem a): Finding the probability that one of the scores in the sample is less than 1411.
Given the mean [tex]\(\mu = 1418\)[/tex] and the standard deviation [tex]\(\sigma = 40\)[/tex], we need to find the probability [tex]\(P(X < 1411)\)[/tex].
1. Calculate the Z-score for an individual score:
The Z-score formula is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
Plugging in the values:
[tex]\[ Z = \frac{1411 - 1418}{40} = \frac{-7}{40} = -0.175 \][/tex]
2. Find the cumulative probability for this Z-score:
Using standard normal distribution tables or a cumulative distribution function (CDF), we find the probability corresponding to [tex]\(Z = -0.175\)[/tex].
[tex]\[ P(Z < -0.175) = 0.4305 \][/tex]
Therefore, the probability that one of the scores in the sample is less than 1411 is:
[tex]\[ P(X < 1411) = 0.4305 \][/tex]
### Problem b): Finding the probability that the average of the scores for the sample of 43 scores is less than 1411.
Given that the sample size [tex]\(n = 43\)[/tex], we need to find the probability [tex]\(P(\bar{X} < 1411)\)[/tex].
1. Calculate the standard error of the mean (SEM):
The standard error of the mean is calculated by:
[tex]\[ SEM = \frac{\sigma}{\sqrt{n}} \][/tex]
Plugging in the values:
[tex]\[ SEM = \frac{40}{\sqrt{43}} \approx \frac{40}{6.5574} \approx 6.099 \][/tex]
2. Calculate the Z-score for the sample mean:
Using the sample mean calculation, the Z-score formula is:
[tex]\[ Z = \frac{\bar{X} - \mu}{SEM} \][/tex]
Plugging in the values:
[tex]\[ Z = \frac{1411 - 1418}{6.099} = \frac{-7}{6.099} \approx -1.1476 \][/tex]
3. Find the cumulative probability for the sample mean's Z-score:
Using standard normal distribution tables or a cumulative distribution function (CDF), we find the probability corresponding to [tex]\(Z = -1.1476\)[/tex].
[tex]\[ P(Z < -1.1476) = 0.1256 \][/tex]
Therefore, the probability that the average of the scores for the sample of 43 scores is less than 1411 is:
[tex]\[ P(\bar{X} < 1411) = 0.1256 \][/tex]
To summarize:
[tex]\[ P(X < 1411) = 0.4305 \][/tex]
[tex]\[ P(\bar{X} < 1411) = 0.1256 \][/tex]
### Problem a): Finding the probability that one of the scores in the sample is less than 1411.
Given the mean [tex]\(\mu = 1418\)[/tex] and the standard deviation [tex]\(\sigma = 40\)[/tex], we need to find the probability [tex]\(P(X < 1411)\)[/tex].
1. Calculate the Z-score for an individual score:
The Z-score formula is:
[tex]\[ Z = \frac{X - \mu}{\sigma} \][/tex]
Plugging in the values:
[tex]\[ Z = \frac{1411 - 1418}{40} = \frac{-7}{40} = -0.175 \][/tex]
2. Find the cumulative probability for this Z-score:
Using standard normal distribution tables or a cumulative distribution function (CDF), we find the probability corresponding to [tex]\(Z = -0.175\)[/tex].
[tex]\[ P(Z < -0.175) = 0.4305 \][/tex]
Therefore, the probability that one of the scores in the sample is less than 1411 is:
[tex]\[ P(X < 1411) = 0.4305 \][/tex]
### Problem b): Finding the probability that the average of the scores for the sample of 43 scores is less than 1411.
Given that the sample size [tex]\(n = 43\)[/tex], we need to find the probability [tex]\(P(\bar{X} < 1411)\)[/tex].
1. Calculate the standard error of the mean (SEM):
The standard error of the mean is calculated by:
[tex]\[ SEM = \frac{\sigma}{\sqrt{n}} \][/tex]
Plugging in the values:
[tex]\[ SEM = \frac{40}{\sqrt{43}} \approx \frac{40}{6.5574} \approx 6.099 \][/tex]
2. Calculate the Z-score for the sample mean:
Using the sample mean calculation, the Z-score formula is:
[tex]\[ Z = \frac{\bar{X} - \mu}{SEM} \][/tex]
Plugging in the values:
[tex]\[ Z = \frac{1411 - 1418}{6.099} = \frac{-7}{6.099} \approx -1.1476 \][/tex]
3. Find the cumulative probability for the sample mean's Z-score:
Using standard normal distribution tables or a cumulative distribution function (CDF), we find the probability corresponding to [tex]\(Z = -1.1476\)[/tex].
[tex]\[ P(Z < -1.1476) = 0.1256 \][/tex]
Therefore, the probability that the average of the scores for the sample of 43 scores is less than 1411 is:
[tex]\[ P(\bar{X} < 1411) = 0.1256 \][/tex]
To summarize:
[tex]\[ P(X < 1411) = 0.4305 \][/tex]
[tex]\[ P(\bar{X} < 1411) = 0.1256 \][/tex]
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