Question

The monthly expenditures on food by single adults living in one neighborhood of Los Angeles are normally distributed with a mean of $370 and a standard deviation of $100. Determine the percentage of samples of size 25 that will have mean monthly expenditures on food within $36 of the population mean expenditure of $370

Answer 1

`P(334<\bar X<406)`

And we can use the following z score formula:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z = (406 -370)/((100)/(√(25)))= 1.8`

`z = (334 -370)/((100)/(√(25)))= 1.8`

And we want thi probability:

`P(-1.8 < Z<1.8)= P(Z<1.8) -P(Z<-1.8) = 0.964-0.036= 0.928`

Step-by-step 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 expenditures of a population, and for this case we know the distribution for X is given by:

`X \sim N(370,100)`

Where
`\mu=370` and
`\sigma=100`

We are interested on this probability

`P(334<\bar X<406)`

And we can use the following z score formula:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

And replacing we got:

`z = (406 -370)/((100)/(√(25)))= 1.8`

`z = (334 -370)/((100)/(√(25)))= 1.8`

And we want thi probability:

`P(-1.8 < Z<1.8)= P(Z<1.8) -P(Z<-1.8) = 0.964-0.036= 0.928`