Dinner for two at The Diner cost on average 45 with a standa

Dinner for two at The Diner cost on average $45 with a standard deviation of $5.

What is the probability a random couple spends between $50 and $58 on dinner at The Diner?

What is the probability a random couple spends MORE than $33 on dinner at The Diner?

What is the probability a random couple spends between $44 and $59 on dinner at The Diner?

let X be the random variable denoting the cost of dinner for two at the Diner.

it is given that X has a average of $45 and a standard deviation of $5

assumption is that X follows a normal distribution.

so X~N(45,5²)

so the probability a random couple spends between $50 and $58 on dinner at The Diner is

P[50<X<58]=P[(50-45)/5<(X-45)/5<(58-45)/5]=P[1<Z<2.6] where Z=(X-45)/5~N(0,1)

=P[Z<2.6]-P[Z<1]=0.995339-0.841345=0.153994 [answer] [using minitab]

so the probability a random couple spends MORE than $33 on dinner at The Diner is

P[X>33]=1-P[X<33]=1-P[(X-45)/5<(33-45)/5]=1-P[Z<-2.4] where Z=(X-45)/5~N(0,1)

=1-0.0081975=0.9918025 [answer] [using minitab]

so the probability a random couple spends between $44 and $59 on dinner at The Diner is

P[44<X<59]=P[(44-45)/5<(X-45)/5<(59-45)/5]=P[-0.2<Z<2.8] where Z=(X-45)/5~N(0,1)

=P[Z<2.8]-P[Z<-0.2]=0.997445-0.420740=0.576705 [answer] [using minitab]