Question

Enter all answers to the nearest tenth. Some of the following numbers will be used more than once and some may not be used. A delivery driver has an average daily gasoline expense of $55.00. The standard deviation is $10.00. The owner takes a sample of 54 bills. What is the probability the mean of his sample will be between $45.00 and $65.00?

Options:
a) 0.8413
b) 0.1587
c) 0.6827
d) 0.3173

Answer 1

To find the probability that the mean of the sample will be between $45.00 and $65.00, we can calculate the z-scores for these values and use a z-table or a calculator to find the probabilities associated with these z-scores. The probability is approximately 0.9954.

Step-by-step explanation:

To find the probability that the mean of the sample will be between $45.00 and $65.00, we can calculate the z-scores for these values using the formula:

z = (x - μ) / (σ / √n)

where x is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size.

Using the given information in the question, we have:

μ = $55.00, σ = $10.00, n = 54

Calculating the z-scores for $45.00 and $65.00:

z1 = ($45.00 - $55.00) / ($10.00 / √54) ≈ -3.06

z2 = ($65.00 - $55.00) / ($10.00 / √54) ≈ 3.06

Next, we can use a z-table or a calculator to find the probabilities associated with these z-scores:

Probability(z1 ≤ Z ≤ z2)

= Probability(Z ≤ 3.06) - Probability(Z ≤ -3.06)

This gives us a probability of approximately 0.9977 - 0.0023 ≈ 0.9954.