Mathematics: A marketing firm is considering making up to three new hires. Given its specific needs, the management feels that there is a 60% chance of hiring at least two candidates. There is only a 6% chance that it will not make any hires and a 16% chance that it will make all three hires. Required: a. What is the probability that the firm will make at least one hire? b. Find the expected value and the standard deviation of the number of hires.

42 : 49= 6 : 7 ( Dividing both sides of the ratio by seven(7))...

A marketing firm is considering making up to three

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08.07.2021

a. 0.94 = 94% probability that the firm will make at least one hire.

b. The expected value of the number of hires is 1.7, and the standard deviation is 0.8062.

Step-by-step explanation:

There is only a 6% chance that it will not make any hires

This means that P(X = 0) = 0.06.

16% chance that it will make all three hires.

This means that P(X = 3) = 0.16

60% chance of hiring at least two candidates.

This means that:

P(X = 2) + P(X = 3) = 0.6

P(X = 2) + 0.16 = 0.6

P(X = 2) = 0.44

Probability of one hire:

The sum of all probabilities, from no hires to three hires, is 1. So

P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 1

0.06 + P(X = 1) + 0.16 + 0.44 = 1

P(X = 1) = 0.66 = 1

P(X = 1) = 0.34

a. What is the probability that the firm will make at least one hire?

$P(X \geq 1) = 1 - P(X = 0) = 1 - 0.06 = 0.94$

0.94 = 94% probability that the firm will make at least one hire.

b. Find the expected value and the standard deviation of the number of hires.

Expected value:

E(X) = 0.06*0 + 0.34*1 + 0.44*2 + 0.16*3 = 1.7

Standard deviation:

$S(X) = \sqrt{0.06*(0-1.7)^2 + 0.34*(1-1.7)^2 + 0.44*(2-1.7)^2 + 0.16*(3-1.7)^2} = 0.8062$

The expected value of the number of hires is 1.7, and the standard deviation is 0.8062.

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