In a recent survey of high school students, it was found that the average amount of money spent on entertainment each week was n
ormally 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) 0.3372
B) 0.0174
C) 0.1628
D) 0.4826
2 answers:
Answer: 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
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