In a recent survey of high school students, it was found that the average amount of money spent on entertainment each week was normally distributed with a mean of $52.30 and a standard deviation of $18.23. assuming these values are representative of all high school students, what is the probability that for a sample of 25, the average amount spent by each student exceeds $60?
a) 0.3372
b) 0.0174
c) 0.1628
d) 0.4826

A

Step-by-step explanation:

To solve this problem, you will first need to calculate the Z value of the variable of interest and then use a Z distribution table to look up the probability. To calculate the Z score, use the following formula.

See attachment

x = the value that is being standardized

m = the mean of the distribution

s = standard deviation of the distribution

Z>$60-$52.30/$18.23

Z>$7.7/$18.23

Z=>4224

then you check the Z value on the standard normal distribution table =.6628

1.0000-0.6628=0.3372

use an equation of ratios to determine the amount of gas needed to fill the tank.

    0.2             0.8

  =    

288.75             x

                                                              231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

                                                                0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal         $3

* = $3465   (answer)

      1           1 gal

hope this you

hit or miss i guess they never miss huh

step-by-step explanation:

you got a boyfriend i bet he doesnt kiss ya