1. First, we must find the total number of students:
N = 0 + 8 + 12 + 4 + 6 + 0 + 1 + 0 + 0 + 0 + 0 = 31
Now, here is the table:
| 0 | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 | |
| Frequency | 0 | 8 | 12 | 4 | 6 | 0 | 1 | 0 | 0 | 0 | 0 |
| Rel. Freq | 0/31= 0.0000 | 8/31 = 0.2581 | 12/31= 0.3871 | 4/31 = 0.1290 | 6/31 = 0.1935 | 0/31 = 0.0000 | 1/31 = 0.0323 | 0/31 = 0.0000 | 0/31 = 0.0000 | 0/31 = 0.0000 | 0/31 = 0.0000 |
2. What is the probability that a household owns:
a. At least two computers?
P(>= 2) = 1 - P(<= 1) = 1 - P(1) - P(0) = 1.0000 - 0.2581 - 0.0000 = 0.7419 = 74.19 %.
b. Exactly two computers?
P(2) = 0.3871 = 38.71 %.
c. Between one and three computers inclusive?
P(between 1 and 3) = P(1) + P(2) + P(3) = 0.2581 + 0.3871 + 0.1290 = 0.7742 = 77.42 %.
d. An odd number of computers?
P(odd number) = P(1) + P(3) + P(5) + P(7) + P(9) = 0.2581 + 0.1290 + 0.0000 + 0.0000 + 0.0000
= 0.3871 = 38.71 %
3. Using the probability distribution in number 1, find the following:
a. Find the mean of the random variable:
mu = mean number of computers = sum (number of computers * probability of each) =
0 * 0.0000 + 1 * 0.2581 + 2 * 0.3871 + 3 * 0.1290 + 4 * 0.1935 + 5 * 0.0000 + 6 * 0.0323 + 7 * 0.0000
+ 8 * 0.0000 + 9 * 0.0000 + 10 * 0.0000 = 0.2581 + 0.7742 + 0.3870 + 0.7740 + 0.1938 = 2.3871
b. Obtain the standard deviation of the random variable.
sigma = standard deviation of the number of computers =
sqrt( sum[ (number of computers - mu)² * (probability of each) ] ) = sqrt[ (0-2.3871)² * 0.0000 +
(1-2.3871)² * 0.2581 + (2-2.3871)² * 0.3871 + (3-2.3871)² * 0.1290 + (4-2.3871)² * 0.1935 +
(5-2.3871)² * 0.0000 + (6-2.3871)² * 0.0323 + (7-2.381)² * 0.0000 + (8-2.381)² * 0.0000 +
(9-2.381)² * 0.0000 + (10-2.381)² * 0.0000 ]
= sqrt[ 0.4966 + 0.0580 + 0.0485 + 0.5034 + 0.4216 ] = sqrt [ 1.5281 ] = 1.2362
c. Draw a probability histogram for the random variable and locate the mean and show one, two, and three standard deviations away from the mean.
Light blue fill = histogram and Blue line = mean
Green line = 1 standard deviation away from the mean
Red line = 2 standard deviations away from the mean
Yellow line = 3 standard deviations away from the mean
d. What is the probability that a randomly selected student will have a household with the number of computers within 2 standard deviations from the mean?
First, we see that within 2 standard deviations from the mean implies that the number of computers must be between mu - 2 * sigma = 2.3871 - 2 * 1.2362 = -0.0853 and mu + 2 * sigma = 2.3871 + 2 * 1.2362 = 4.8595, or, in other words, between 0 and 4 inclusive, so now we get:
P(2 sigmas from the mean) = P(between 0 and 4) = 1 - P(5) - P(6) - P(between 7 and 10)
= 1.0000 - 0.0000 - 0.0323 - 0.0000 = 0.9677 = 96.77 %.
