Respuesta :
Answer:
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 [tex]\bar X[/tex] = sample mean
Now, the z score probability distribution for sample mean is given by;
Z = [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] ~ N(0,1)
where, [tex]\mu[/tex] = average daily balance of its customers = $413
[tex]\sigma[/tex] = 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 < [tex]\bar X[/tex] < $410.4) = P([tex]\bar X[/tex] < $410.4) - P([tex]\bar X[/tex] [tex]\leq[/tex] 402.6)
P([tex]\bar X[/tex] < $410.4) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] < [tex]\frac{410.4-413}{\frac{68}{\sqrt{61} } }[/tex] ) = P(Z < -0.29) = 1 - P(Z [tex]\leq[/tex] 0.29)
= 1 - 0.61409 = 0.38591
P([tex]\bar X[/tex] [tex]\leq[/tex] $402.6) = P( [tex]\frac{\bar X-\mu}{\frac{\sigma}{\sqrt{n} } }[/tex] [tex]\leq[/tex] [tex]\frac{402.6-413}{\frac{68}{\sqrt{61} } }[/tex] ) = P(Z [tex]\leq[/tex] -1.19) = 1 - P(Z < 1.19)
= 1 - 0.88298 = 0.11702
Therefore, P($402.6 < [tex]\bar X[/tex] < $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.
