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?

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

Step-by-step 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 = sample mean

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

         Z = ~ N(0,1)

where, = average daily balance of its customers = $413

            = 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 < < $410.4) = P( < $410.4) - P( 402.6)

P( < $410.4) = P( < ) = P(Z < -0.29) = 1 - P(Z 0.29)

                                                         = 1 - 0.61409 = 0.38591

 P( $402.6) = P( ) = P(Z -1.19) = 1 - P(Z < 1.19)

                                                           = 1 - 0.88298 = 0.11702

Therefore, P($402.6 < < $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.

b) 36

step-by-step explanation:

18 times 2 is 36

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