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

Answer:

$P(38.6$

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

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

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

$P(-0.4$

Step-by-step 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:

$P(38.6$

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

$z=\frac{X -\mu}{\sigma}$

And replacing we got:

$z=\frac{38.6- 45}{16}= -0.4$

$z=\frac{57.8- 45}{16}= 0.8$

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

$P(-0.4$