Question

A bank has kept records of the checking balances of its customers and determined that the average daily balance of its customers is $413 with a standard deviation of $68. A random sample of 61 checking accounts is selected. What is the probability that the sample mean will be between $402.6 and $410.4?

Answer 1

Probability that the sample mean will be between $402.6 and $410.4 is 0.2689.

Explanation:

We are given that a bank has kept records of the checking balances of its customers and determined that the average daily balance of its customers is $413 with a standard deviation of $68.

A random sample of 61 checking accounts is selected.

Let
`\bar X` = sample mean

Now, the z score probability distribution for sample mean is given by;

Z =
`(\bar X-\mu)/((\sigma)/(√(n) ) )` ~ N(0,1)

where,
`\mu` = average daily balance of its customers = $413

`\sigma` = standard deviation = $68

n = sample of checking accounts = 61

So, probability that the sample mean will be between $402.6 and $410.4 is given by = P($402.6 <
`\bar X` < $410.4) = P(
`\bar X` < $410.4) - P(
`\bar X`
`\leq` 402.6)

P(
`\bar X` < $410.4) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )` <
`(410.4-413)/((68)/(√(61) ) )` ) = P(Z < -0.29) = 1 - P(Z
`\leq` 0.29)

= 1 - 0.61409 = 0.38591

P(
`\bar X`
`\leq` $402.6) = P(
`(\bar X-\mu)/((\sigma)/(√(n) ) )`
`\leq`
`(402.6-413)/((68)/(√(61) ) )` ) = P(Z
`\leq` -1.19) = 1 - P(Z < 1.19)

= 1 - 0.88298 = 0.11702

Therefore, P($402.6 <
`\bar X` < $410.4) = 0.38591 - 0.11702 = 0.2689

Hence, probability that the sample mean will be between $402.6 and $410.4 is 0.2689.