Question in a recent survey of high school students, it is found that the average amount of money spend on entertainment each week was normally distributed with a mean of $52.30 and a standard deviation of $18.23 assuming that these value are representative of all high school students, what is the probability that for a sample of 25 the average amount spend by each student exceeds $60.00
Answer Let the average amount of money spend on entertainment each week foloes normal distribution with a mean of $52.30 and a standard deviation of $18.23 assuming that these value are representative of all high school students, To find the probability that for a sample of 25 the average amount spend by each student exceeds $60.00
P(X>60) =P(X-mean/sd > (60-52.30)/18.23)
= P(Z>0.4262) Bothsides subtract with mean and devide by Sd,by defn z=x=mean/sd
=1-P(Z<0.4262)
P(Z<0.4262) is obtained with help of EXCEL function =NORMDIST(60,52.3,18.23,TRUE)
