Question

A bank has kept records of the checking balances of its customers and determined that the average daily balance of its customers is $300 with a standard deviation of $56. A random sample of 200 checking accounts is selected. You are interested in calculating the following probabilities below.1. Assuming that the population of the checking account balances is normally distributed, what is the probability that a randomly selected account has a balance of more than $305?2. What is the probability that the mean balance for the selected sample is above $295?3. What is the probability that the mean balance for the selected sample is below $290?4. What is the probability that the mean balance for the selected sample is between $302 and $304?

Answer 1

Answer:

1. P(X > 305) = $0.1038

2. P ( X > 295) = $0.8962

3. P ( X > 290) = $0.0057

4. P(302 < X < 304 ) = $0.1488

Explanation:

Solution:

Data Given:

Mean = u = $300

SD = Standard Deviation = $56

Sample Size = n = 200

uX = u = 300

SDX = $\frac{SD}{\sqrt{n} }$ = $\frac{56}{\sqrt{200} }$ = 3.96

1.

P(X > 305) = 1-P ($\frac{X - uX}{SDX} < \frac{305 - 300}{3.96}$)

P(X > 305) = 1-P (Z < 1.26)

Using Standard Normal Table, we have:

P(X > 305) = 1 - 0.8962

Probability = $0.1038

2.

P ( X > 295) = 1 - P ( $\frac{X - uX }{SDX} < \frac{295 - 300}{3.96}$ )

P ( X > 295) = 1 - P (Z< 1.26)

Using standard normal table, we have:

P ( X > 295) = 1 - 0.1038

P ( X > 295) = $0.8962

3.

P ( X > 290) =  P ( $\frac{X - uX }{SDX} < \frac{290 - 300}{3.96}$ )

P ( X > 290) = P ( z< -2.53)

Using Standard normal table, we have:

P ( X > 290) = $0.0057

4.

P(302 < X < 304 ) = P ( $\frac{302 - 300}{3.96} < \frac{X - uX}{SDX} < \frac{304 - 300}{3.96}$ )

P(302 < X < 304 ) = P ( 0.51 < z < 1.01)

P(302 < X < 304 ) = P (z < 1.01) - P (z < 0.51)

P(302 < X < 304 ) = 0.8438 - 0.6950

P(302 < X < 304 ) = $0.1488