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blondinia [14]
3 years ago
5

Suppose a large labor union wishes to estimate the mean number of hours per month a union member is absent from work. The union

decides to sammple 348 of its members at random and monitor the working time of each of them for 1 month. At the end of the month the total number of hours absent from work is recorded for each employee . if the mean and standard deviation of the sample are x = 7.5 hours and s = 3.5 hours, find a 99% confidence interval for the true mean number of hours a union member is absent per month. Round to the nearest thousandth.
A) 7.5 ± .258
B) 7.5 ± .186
C) 7.5 ± .026
D) 7.5 ± .483
Mathematics
1 answer:
viva [34]3 years ago
3 0

Answer:

c

Step-by-step explanation:

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16. A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours
Reika [66]

Answer:

a) 23.11% probability of making exactly four sales.

b) 1.38% probability of making no sales.

c) 16.78% probability of making exactly two sales.

d) The mean number of sales in the two-hour period is 3.6.

Step-by-step explanation:

For each phone call, there are only two possible outcomes. Either a sale is made, or it is not. The probability of a sale being made in a call is independent from other calls. So we use the binomial probability distribution to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

C_{n,x} = \frac{n!}{x!(n-x)!}

And p is the probability of X happening.

A telemarketer makes six phone calls per hour and is able to make a sale on 30% of these contacts. During the next two hours, find:

Six calls per hour, 2 hours. So

n = 2*6 = 12

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a. The probability of making exactly four sales.

This is P(X = 4).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 4) = C_{12,4}.(0.3)^{4}.(0.7)^{8} = 0.2311

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b. The probability of making no sales.

This is P(X = 0).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 0) = C_{12,0}.(0.3)^{0}.(0.7)^{12} = 0.0138

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c. The probability of making exactly two sales.

This is P(X = 2).

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

P(X = 2) = C_{12,2}.(0.3)^{2}.(0.7)^{10} = 0.1678

16.78% probability of making exactly two sales.

d. The mean number of sales in the two-hour period.

The mean of the binomia distribution is

E(X) = np

So

E(X) = 12*0.3 = 3.6

The mean number of sales in the two-hour period is 3.6.

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Answer:

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