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

2 and 1.

step-by-step explanation:

plug in each answer. 0 doesn't work, 1 does. so does 2. 4 doesn't:

0 = 1 is not true.

2 = 2 is true.

4 = 4 is true.

8 = 16 is not true.

hope this !

the discriminant is 64 + 4*1*12 = 112

this is prime.

step-by-step explanation: