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Privacy and information sharing. Refer to the Pew Internet & American Life Project Survey (January 2016), Exercise 4.48 (p. 239). The survey revealed that half of all U.S. adults would agree to participate in a free cost-saving loyalty card program at a grocery store, even if the store could potentially sell these data on the customer’s shopping habits to third parties. In a random sample of 250 U.S. adults, let x be the number who would participate in the free loyalty card program.

a. Find the mean of x. (This value should agree with your answer to Exercise 4.48c.)

b. Find the standard deviation of x.

c. Find the z-score for the value x = 200.

d. Find the approximate probability that the number of the 250 adults who would participate in the free loyalty card program is less than or equal to 200.

Step by step solution

01

Given information

Referring to the Paw Internet & American Life Project Survey (January 2016), exercise 4.48.

x be the participate in the free loyalty card program, which follows a binomial distribution with n = 250 and p = 0.5

02

Calculation of mean

$$\begin{gathered} a. \\ \mathrm{μ}=np \\ =250×0.5 \\ =125 \\ \mathrm{μ}=125 \end{gathered}$$

Therefore, the mean of x is 125.

03

Calculation of standard deviation

$$\begin{gathered} b. \\ \mathrm{σ}=\sqrt{np(1-p)} \\ =\sqrt{250×0.5×(1-0.5)} \\ =\sqrt{250×0.5×0.5} \\ =7.9057 \\ \mathrm{σ}=7.9057 \end{gathered}$$

Therefore, the variance of x is 7.9057.

04

Calculation of z-score

$$\begin{gathered} c. \\ \mathrm{μ}=125 \\ \mathrm{σ}=7.9057 \\ x=200 \end{gathered}$$

The z-score is,

$$\begin{gathered} z=\frac{x-\mathrm{μ}}{\mathrm{σ}} \\ =\frac{200-125}{7.9057} \\ =9.4868 \\ z=9.4868 \end{gathered}$$

Therefore, the z-score is 9.4868.

05

Finding the approximate probability

$$\begin{gathered} d. \\ P(x≤200)=P(x≤200) \\ =P(z≤9.4868) \\ =1 \\ P(x≤200)=1 \end{gathered}$$

Therefore, the approximate probability is 1.

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