At a large publishing company, the mean age of proofreaders is 36.2 years, and the standard deviation is 3.7 years. If a random sample of 15 proofreaders is selected, find the probability that the mean age of the proofreaders in the sample will be between 36 and 37.5 years.Assume the variable is normally distributed.
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To find the probability that the mean age of the proofreaders in the sample will be between 36 and 37.5 years, we need to calculate the z-scores for both values and then find the corresponding areas under the normal curve.
Step 1: Calculate the z-score for the lower limit (36):
The formula to calculate the z-score is:
z = (x - μ) / (σ / √n)
where:
x = the lower limit of 36
μ = the mean age of the proofreaders (36.2 years)
σ = the standard deviation (3.7 years)
n = the sample size of 15
z1 = (36 - 36.2) / (3.7 / √15)
Simplifying:
z1 = (-0.2) / (3.7 / √15)
Calculating the value:
z1 ≈ -0.2 / 0.955
z1 ≈ -0.2094
Step 2: Calculate the z-score for the upper limit (37.5):
Using the same formula:
z2 = (37.5 - 36.2) / (3.7 / √15)
Simplifying:
z2 = (1.3) / (3.7 / √15)
Calculating the value:
z2 ≈ 1.3 / 0.955
z2 ≈ 1.3608
Step 3: Calculate the probability between the two z-scores:
To find the probability, we need to use a standard normal distribution table or a calculator. We want to find the area to the left of z2 and subtract the area to the left of z1.
P(36 ≤ x ≤ 37.5) = P(z1 ≤ z ≤ z2)
Using a standard normal distribution table or a calculator, you can find the corresponding probabilities for z1 and z2:
P(z1 ≤ -0.2094) ≈ 0.4182
P(z ≤ 1.3608) ≈ 0.9115
Subtracting the two probabilities:
P(36 ≤ x ≤ 37.5) ≈ 0.9115 - 0.4182
P(36 ≤ x ≤ 37.5) ≈ 0.4933
Therefore, the probability that the mean age of the proofreaders in the sample will be between 36 and 37.5 years is approximately 0.4933.