Question

According to Crown ATM Network, the mean ATM withdrawal is $67. Assume that the standard deviation for withdrawals is $35. If a random sample of 50 ATM withdrawals is obtained, determine the probability of obtaining a sample mean withdrawal amount between $70 and $75.

Answer 1

`z = (70-67)/((35)/(√(50)))= 0.606`

`z = (75-67)/((35)/(√(50)))= 1.616`

And using a calculator, excel or the normal standard table we have that:

`P(0.606<Z<1.616)=P(z<1.616) -P(Z<0.606)= 0.947-0.728= 0.219`Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Solution to the problem

Let X the random variable that represent the withdrawals of a population, and for this case we know the following info

`\mu=67` and
`\sigma=35`

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean
`\bar X` is given by:

`\bar X \sim N(\mu, (\sigma)/(√(n)))`

We can find the probability of interest like this:

`z = (70-67)/((35)/(√(50)))= 0.606`

`z = (75-67)/((35)/(√(50)))= 1.616`

And using a calculator, excel or the normal standard table we have that:

`P(0.606<Z<1.616)=P(z<1.616) -P(Z<0.606)= 0.947-0.728= 0.219`