Question

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 represented of all high school students, what is the probability that for a sample of 25, the average amount sent by each student exceeds $60?
a)0.3372
b)0.1628
c)0.4826
d)0.0174

Answer 1

Solution :

Given that ,

mean = =  $52.30

standard deviation = = $18.23

n = 25

= $52.30 and

= / n = 18.23/ 25 = 3.646

P( >60 ) = 1 - P( < 60)

= 1 - P(( - ) / < (60- 52.30) / 3.646)

= 1 - P(z < 2.11)

    Using standard normal table,

P( >60) = 1- 0.9826

Probability = 0.0174

d)0.0174