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A department store has determined that 25% of all their sales are credit sales.  A random sample of 60 sales is selected.  The probability that the sample proportion will be less than 0.3 is                           .

Question

A department store has determined that 25% of all their sales are credit sales.  A random sample of 60 sales is selected.  The probability that the sample proportion will be less than 0.3 is                           .

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Solution

To solve this problem, we will use the concept of Normal Approximation to the Binomial Distribution.

Step 1: Identify the given values. The proportion of credit sales, p = 0.25 (25%) The sample size, n = 60 We want to find the probability that the sample proportion is less than 0.3.

Step 2: Calculate the mean and standard deviation. The mean (μ) of a binomial distribution is np. So, μ = np = 60 * 0.25 = 15 The standard deviation (σ) is sqrt(np(1-p)). So, σ = sqrt(60 * 0.25 * (1 - 0.25)) = sqrt(11.25) = 3.35 (approximately)

Step 3: Standardize the desired proportion. We want to find the probability that the sample proportion (p̂) is less than 0.3. In terms of the number of sales, this is 0.3 * 60 = 18 sales. To standardize this, we subtract the mean and divide by the standard deviation. This gives us a z-score. Z = (18 - 15) / 3.35 = 0.895 (approximately)

Step 4: Look up the probability associated with this z-score. Using a standard normal distribution table or a calculator, we find that the probability associated with a z-score of 0.895 is approximately 0.815.

So, the probability that the sample proportion will be less than 0.3 is approximately 0.815.

This problem has been solved

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