Question

Fill in the missing information. Tim Worker is doing his budget. He discovers that the average miscellaneous expense is $45.00 with a standard deviation of $16.00. What percent of his expense in this category would he expect to fall between $38.60 and $57.80?

Answer 1

`P(38.6 <X <57.8)`

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

`z=(X -\mu)/(\sigma)`

And replacing we got:

`z=(38.6- 45)/(16)= -0.4`

`z=(57.8- 45)/(16)= 0.8`

We can find the probability of interest using the normal standard table and with the following difference:

`P(-0.4 <z<0.8)= P(z<0.8) -P(z<-0.4) = 0.788-0.345= 0.443`

Explanation:

Let X the random variable who represent the expense and we assume the following parameters:

`\mu = 45, \sigma 16`

And for this case we want to find the percent of his expense between 38.6 and 57.8 so we want this probability:

`P(38.6 <X <57.8)`

And we can assume a normal distribution and then we can solve the problem with the z score formula given by:

`z=(X -\mu)/(\sigma)`

And replacing we got:

`z=(38.6- 45)/(16)= -0.4`

`z=(57.8- 45)/(16)= 0.8`

We can find the probability of interest using the normal standard table and with the following difference:

`P(-0.4 <z<0.8)= P(z<0.8) -P(z<-0.4) = 0.788-0.345= 0.443`