Question

According to Crown ATM Network, the mean ATM withdrawal is $67. Assume thatthe standard deviation for withdrawals is $35. If a randomly sample of 50 ATMwithdrawals is obtained, what is the probability of obtaining a sample meanwithdrawal amount between $70 and $75, rounded to the nearest ten-thousandth (4decimal places)?

Answer 1

We are given the following information

Mean ATM withdrawal = μ = $67

Standard deviation of ATM withdrawal = σ = $35

Sample size = n = 50

The probability of obtaining a sample mean withdrawal amount between $70 and $75 is given by

`\begin{gathered} P(70\le\bar{x}\le75)=P(\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}}\le z\le\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt[]{n}}}) \\ P(70\le\bar{x}\le75)=P(\frac{70-67}{\frac{35}{\sqrt[]{50}}}\le z\le\frac{75-67}{\frac{35}{\sqrt[]{50}}}) \end{gathered}`