Let X denote the outstanding balances of customers of a firm. From past experiences, X is well approximated by a normal distribution with a mean of 44 and a variance of 100. If an auditor takes a random sample of 36 accounts what is the probability that the mean balance will be less than 44? (Your answer should be correct to one decimal place.)
Question
Let X denote the outstanding balances of customers of a firm. From past experiences, X is well approximated by a normal distribution with a mean of 44 and a variance of 100. If an auditor takes a random sample of 36 accounts what is the probability that the mean balance will be less than 44? (Your answer should be correct to one decimal place.)
Solution
To solve this problem, we need to use the concept of the standard normal distribution (Z-distribution).
Step 1: Identify the given values. The mean (μ) = 44 The variance (σ^2) = 100, so the standard deviation (σ) = sqrt(100) = 10 The sample size (n) = 36 The value we are comparing to (X) = 44
Step 2: Calculate the standard error. The standard error (SE) = σ/sqrt(n) = 10/sqrt(36) = 10/6 = 1.67
Step 3: Calculate the Z-score. The Z-score = (X - μ) / SE = (44 - 44) / 1.67 = 0
Step 4: Look up the Z-score in a standard normal distribution table or use a calculator with a normal distribution function to find the probability. The probability that Z is less than 0 is 0.5 (or 50%) because a standard normal distribution is symmetric about 0, with half the distribution below 0 and half above.
So, the probability that the mean balance will be less than 44 is 0.5 or 50%.
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