The manager of a video store has compiled the following table, which gives the probabilities that a customer will buy 0, 1, 2, 3, or 4 dvds. how many dvds can a given customer be expected to buy? dvds - 0 1 2 3 4 probability - .42 .36 .14 .05 .03

0.91 DVDs

Step-by-step explanation:

The expected number of DVDs is the sum of products of the number of DVDs and their probability:

  E(n) = 0·0.42 + 1·0.36 + 2·0.14 + 3·0.05 + 4·0.03

  = 0 + 0.36 + 0.28 + 0.15 + 0.12

  = 0.91

A given customer can be expected to buy 0.91 DVDs.

0.91 dvds

It’s correct on acellus

Step-by-step explanation

from the figure we get,

the x axis shows   the months and,

the y-axis shows the number of satisfied customer.

each line or each cell in upward direction (y-axis) shows 500.

the appropriate scale for the y - axis is: -

a = 2000 customers

b = 4000 customers

c = 6000 customers

d = 8000 customers

e = 10000 customers

f = 12000 customers

g = 14000 customers

b) 1/3 * 1/5

step-by-step explanation:

each smallest part is 1/15 since the written one is 1/5 and it's divided by 3 equal parts. (1/5 )/3 = 1/15 = 1/3 * 1/5