Answer:

the mean: 9.04

the variance : 3.0184

standard deviation: 1.7374

the probability that the number of rides for a randomly selected Disney World visitor is within 1 standard deviation of the mean: 0.84

the probability that the number of rides for a randomly selected Disney World visitor is less than 1 standard deviation above the mean:0.95

Step-by-step explanation

First, we determine the average number of attractions for a Disney World visitor

?E(x)=∑xp(x)=5∗0.04+6∗0.07+7∗0.09+8∗0.12+9∗0.20+10∗0.30+11∗0.13+12∗0.05=? 9.04

 second, we determine the variance and the deviation

?V(x)=E(x2)−(E(x))2?

?E(x2)=∑x2p(x)? =?52∗0.04+62∗0.07+72∗0.09+82∗0.12+92∗0.20+102∗0.30+112∗0.13+122∗0.05=? 84.74

?V(x)=84.74−(9.04)2=3.0184?

?σ=3.0184?=1.7374?

Third, we determine the probability that the number of trips for a randomly selected Disney World visitor is within 1 standard deviation of the mean.

?p(9.04−1.7374<x<9.04+1.7374)=P(7.3026x<10.7774)=? ?p(7<x<11)=0.09+0.12+0.20+0.30+0.13? =0.84

 fourth, we determine the probability that the number of trips for a randomly selected Disney World visitor is less than 1 standard deviation above the mean

?p(x≤11)=0.04+0.07+0.09+0.12+0.20+0.30+0.13=0.95?