To find the probability that the sample mean will be within 6 units of the mean, we'll need to use the concept of the standard error.
The standard error, denoted as SE, measures the average amount that the sample mean differs from the population mean. It is calculated using the formula:
SE = σ / √n
where σ is the population standard deviation, and n is the sample size.
In this case, the population standard deviation (σ) is given as 15, and the sample size (n) is 26. Plugging these values into the formula, we can calculate the standard error:
SE = 15 / √26 ≈ 2.94 (rounded to two decimal places)
Now, to find the probability that the sample mean will be within 6 units of the mean, we need to find the corresponding z-scores for the lower and upper limits.
Lower limit z-score:
Z = (X - μ) / SE
Z = (40 - 46) / 2.94 ≈ -2.04
Upper limit z-score:
Z = (X - μ) / SE
Z = (55 - 46) / 2.94 ≈ 3.06
To find the probability between these z-scores, we can use a standard normal distribution table or a calculator.
From the standard normal distribution table or calculator, the probability corresponding to a z-score of -2.04 is approximately 0.0207, and the probability corresponding to a z-score of 3.06 is approximately 0.9782.
To find the probability between the two limits, we subtract the lower probability from 1 (to include the area up to the lower limit) and then subtract the upper probability from the result:
P(40 ≤ X ≤ 55) ≈ 1 - 0.0207 - 0.9782 ≈ 0.0011
Therefore, the probability that the sample mean will be within 6 units of the mean is approximately 0.0011 (correct to four decimal places).