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1. Consider the probability distribution;
×
10 15 18 20 25 33
X) 0.09 0.12 0.38
P (x)
0.18 0.12 0.11
2. The teacher wants to know the average student who are la
class in a day. Let x be the number of late. Below is her data.
X 1 2 3 4 5
P (x)
0.1
0.26 0.3 0.04
0.3
3. The number of policy that an financial advisor sells over a o
period has the following probability distribution.
5
☑0 1 2 3 4
P(x) 0.05 0.30 0.25 0.20 0.15 0.05
4. In a multinational corporation, a Human Resources manager
to be sure that they do not show racial discrimination in hiring e
ees and consultants. Among 500 employees, there are 286 Fili
78 Koreans(2), 100 Americans(3), and 36 Chinese(4). Let x be
Nationality.
5. A bank president feels that each savings account customer h
average, two credit cards. The following distribution represents
ber of
credit cards people own. Is the president correct?
X
P(x) 0.17 0.32 0.38 0.08 0.05
01234
give the answer ​

Sagot :

Yankeriiiidn

1. The average number of late students in a day is calculated by multiplying each number of late students by its probability, then summing up these products. Therefore, the calculation is as follows:

Average = (1*0.1) + (2*0.26) + (3*0.3) + (4*0.04) + (5*0.3)

Average = 0.1 + 0.52 + 0.9 + 0.16 + 1.5

Average = 3.18

Therefore, the average number of late students in a day is 3.18.

2. The probability distribution of the number of credit cards people own does not have a mean of 2. To find the mean, we multiply each number of credit cards by its probability and sum these products. The calculation is as follows:

Mean = (0*0.17) + (1*0.32) + (2*0.38) + (3*0.08) + (4*0.05)

Mean = 0 + 0.32 + 0.76 + 0.24 + 0.2

Mean = 1.52

Thus, the mean number of credit cards people own based on the given distribution is 1.52, not 2 as assumed by the bank president.

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