00:01
Today, us box office ticket prices are on average 7 .50 with variance approximately 4 .41, the distribution appears to be normal.
00:13
Here we have to find what is the probability a ticket will cost more than $112 .50.
00:26
And in part b we have to determine probability that the ticket price is between dollar 6 .50 and dollar 7 .50 and part c we have to find 60 % of the tickets cost above what price and in last we have to determine at the 8th percentile of this distribution so the solution for this question is as follows here let x.
01:19
Defines us box office ticket price here it is given that x is normally distributed at the 7 .50 comma 4 .41 so the value of jad is x minus 7 .50 divided by 2 .1 now we have to solve question part a here in part a we have to find the probability that a ticket will cost more than dollar 12 .50.
02:16
So px greater than 12 .50 is equal to 1 minus p x less than 12 .50.
02:30
This is equal to 1 minus p x minus 7 .50 divided by 2 .1 .1 less then equal to 12 .50 minus 7 .50 divided by 2 .1.
02:52
This is equal to 1 minus p z less than equal to 2 .3 810.
03:01
So this is equal to 0 .0086.
03:06
Hence the required answer is 0 .0086.
03:10
Now we have to solve question power...