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Answer :
The probability that a sample of size n=25 is randomly selected with a mean less than 97.9 is 0.7852.
To find the probability that a single randomly selected value is less than 97.9, we need to calculate the z-score and use the standard normal distribution table.
First, we calculate the z-score using the formula: z = (x - μ) / σ
Where x is the value we want to find the probability for, μ is the mean, and σ is the standard deviation.
Plugging in the values, we get:
z = (97.9 - 96.6) / 8.2
z = 0.15 / 8.2
z ≈ 0.0183
Next, we use the standard normal distribution table or a calculator to find the corresponding probability. Looking up the z-score of 0.0183 in the table, we find the probability is approximately 0.5083.
Therefore, the probability that a single randomly selected value is less than 97.9 is 0.5083 (rounded to 4 decimal places).
Now, let's find the probability that a sample of size n=25 is randomly selected with a mean less than 97.9. This is the same as finding the probability that the sample mean (M) is less than 97.9.
Since we are dealing with the sample mean, we need to use the formula for the standard error of the mean (SE):
SE = σ / √n
Where σ is the standard deviation and n is the sample size.
Plugging in the values, we get:
SE = 8.2 / √25
SE = 8.2 / 5
SE = 1.64
Now, we calculate the z-score using the formula: z = (x - μ) / SE
Where x is the value we want to find the probability for, μ is the mean, and SE is the standard error of the mean.
Plugging in the values, we get:
z = (97.9 - 96.6) / 1.64
z = 1.3 / 1.64
z ≈ 0.7927
Using the standard normal distribution table or a calculator, we find that the probability corresponding to a z-score of 0.7927 is approximately 0.7852.
Therefore, the probability that a sample of size n=25 is randomly selected with a mean less than 97.9 is 0.7852 (rounded to 4 decimal places).
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