Question

You may need to use the appropriate appendix table or technology to answer this question. A bank has kept records of the checking balances of its customers and determined that the average dally balance of its customers is $300 with a standard deviation of s48. A random sample of 144 checking accounts is selected. (Round your answers to four decimal places.)

(a) What is the probability that the sample mean will be more than $291.80 ?
(b) What is the probability that the sample mean wil be less than $309 ?
(c) What is the probability that the sample mean will be between $303 and $306 ?
(d) What is the probability that the sample mean will be at least $295 ?

Answer 1

To answer the given questions about the probability of sample means, we need to calculate z-scores and use the standard normal distribution table. The probabilities are as follows: (a) practically 0, (b) practically 1, (c) approximately 0.067, and (d) approximately 0.006.

Step-by-step explanation:

To answer the given questions, we need to use the concept of sampling distribution and the Central Limit Theorem.

(a) To find the probability that the sample mean will be more than $291.80, we need to calculate the z-score and find the area under the normal curve. The formula for calculating the z-score is: 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. Plugging in the values, we get: z = ($291.80 - $300) / ($48 / √144) = -8.20

Using a standard normal distribution table, we can find that the area to the right of z = -8.20 is practically 0. Therefore, the probability that the sample mean will be more than $291.80 is practically 0.

(b) To find the probability that the sample mean will be less than $309, we use the same formula to calculate the z-score: z = ($309 - $300) / ($48 / √144) = 5.40. Using the standard normal distribution table again, we find that the area to the left of z = 5.40 is practically 1. Therefore, the probability that the sample mean will be less than $309 is practically 1.

(c) To find the probability that the sample mean will be between $303 and $306, we need to calculate the z-scores for both values: z1 = ($303 - $300) / ($48 / √144) = 1.50, and z2 = ($306 - $300) / ($48 / √144) = 3.00. Using the standard normal distribution table, we find that the area between z = 1.50 and z = 3.00 is approximately 0.067. Therefore, the probability that the sample mean will be between $303 and $306 is approximately 0.067.

(d) To find the probability that the sample mean will be at least $295, we calculate the z-score: z = ($295 - $300) / ($48 / √144) = -2.50. Using the standard normal distribution table, we find that the area to the right of z = -2.50 is approximately 0.006. Therefore, the probability that the sample mean will be at least $295 is approximately 0.006.