Question

When an opinion poll selects cell phone numbers at random to dial, the cell phone exchange is first selected and then random digits are added to form a complete telephone number. When using this procedure to generate random cell phone numbers, approximately 55% of the cell numbers generated correspond to working numbers. You watch a pollster dial 15 cell numbers that have been selected in this manner.
(a) What is the mean number of calls that reach a working cell number?
0.55
15
8.25
30
(b) What is the standard deviation ????σ of the count of calls that reach a working cell number?
0.55
2.872
1.927
3.7125
(c) Suppose that the probability of reaching a working cell number was p=0.70p=0.70 . How does this new pp affect the standard deviation?
When p=0.70,σ=3.24 calls.
When p=0.70,σ=3.15 calls.
When p=0.70,σ=1.775 calls.
When p=0.70,σ=10.5 calls.
What would be the standard deviation if p=0.80 ? (Enter your answer rounded to three decimal places.)
σ=
What does your work show about the behavior of the standard deviation of a binomial distribution as the probability of a success gets closer to 1 ?
As p approaches 1 , the standard deviation increases (that is, it approaches 0).
As p approaches 1 , the standard deviation increases (that is, it gets further and further from 0)
As p approaches 1 , the standard deviation remains constant.
As p approaches 1 , the standard deviation decreases (that is, it approaches 0).

Answer 1

Answer:

a. $Mean = 8.25$

b. $SD = 3.7125$

c. $SD = 3.15$

d. $SD = 2.4$

e. As p approaches 1 , the standard deviation increases (that is, it approaches 0).

Step-by-step explanation:

Solving (a):

$p = 55\%$

$n = 15$

Find Mean

Mean is calculated as thus:

$Mean = n * p$

$Mean = 15 * 55\%$

Convert percentage to decimal

$Mean = 15 * 0.55$

$Mean = 8.25$

Solving (b):

Calculate Standard Deviation (SD)

Standard deviation is calculated as thus:

$SD = n * p * (1 - p)$

$SD = 15 * 55\% * (1 - 55\%)$

Convert percentage to decimal

$SD = 15 * 0.55 * (1 - 0.55)$

$SD = 15 * 0.55 * 0.45$

$SD = 3.7125$

Solving (c):

Calculate Standard Deviation if p = 0.7

Using same formula used in (b) above

$SD = n * p * (1 - p)$

$SD = 15 * 0.7 * (1 - 0.7)$

$SD = 15 * 0.7 * 0.3$

$SD = 3.15$

Solving (d):

Calculate Standard Deviation if p = 0.8

Using same formula used in (b) & (c) above

$SD = n * p * (1 - p)$

$SD = 15 * 0.8 * (1 - 0.8)$

$SD = 15 * 0.8 * 0.2$

$SD = 2.4$

Solving (e):

What does the working show

In (b)

When p = 0.5

SD = 3.7125

In (c)

When p = 0.7

SD = 3.15

In (d)

When p = 0.8

SD = 2.4

Notice that as the value of p increases, the standard deviation gets closer to 0.