Akron cinema sells an average of 500 tickets on mondays, with a standard deviation of 50 tickets. if a simple random sample is taken of the mean amount of ticket sales from 30 mondays in a year, what is the probability that the mean will be greater than 510?

Step-by-step explanation:

Assuming the number of tickets sales from Mondays is normally distributed. the formula for normal distribution would be applied. It is expressed as

z = (x - u)/s

Where

x = ticket sales from monday

u = mean amount of ticket

s = standard deviation

From the information given,

u = 500 tickets

s = 50 tickets

We want to find the probability that the mean will be greater than 510. It is expressed as

P(x greater than 510) = 1 - P(x lesser than or equal to 510)

For x = 510

z = (510 - 500)/50 = 0.2

Looking at the normal distribution table, the probability corresponding to the z score is 0.9773

P(x greater than 510) = 1 - 0.9773 = 0.0227

The third set  49,17,4,18.

Step-by-step explanation:

Mean for the first set is (26+6+20+36)/4 = 88/4 = 22.

Mean for the 2nd set = 28+15+30+15 / 4 = 22 also.

Mean for the 3rd set = 49+17+4+18 / 4 = 22

Mean for the 4th set =  14+20+24+30/4 ==22.

The mean is the same for all 4 data sets.

On inspection, the set which has the biggest values furthest from 22 is the 3rd set ( values 4 and 49.) so this looks like the one.

The other with more spread than the rest is the first set . We can confirm by working out the Mean Absolute deviation for both sets:

MAD for set 1 =[ (26-22) + (22-6) + (22-20) + (36-22)] / 4  = 9.

MAD for set 3 = [(49-22) + (22-17) + (22-4) + (22-18) ]  / 4  =  13.5.

So it is the 3rd set.