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Q104E

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Statistics For Business And Economics

Found in: Page 263

Book edition 13th

Author(s) James T. McClave, P. George Benson, Terry Sincich

Pages 888 pages

ISBN 9780134506593

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Short Answer

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.

$a . μ = 125 b . σ = 7.9057 c . z = 9.4868 d . The approximate probability is 1$

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: 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

Step 2: Calculation of mean

$a . μ = n p = 250 \times 0.5 = 125 μ = 125$

Therefore, the mean of x is 125.

Step 3: Calculation of standard deviation

$b . σ = \sqrt{n p (1 - p)} = \sqrt{250 \times 0.5 \times (1 - 0.5)} = \sqrt{250 \times 0.5 \times 0.5} = 7.9057 σ = 7.9057$

Therefore, the variance of x is 7.9057.

Step 4: Calculation of z-score

$c . μ = 125 σ = 7.9057 x = 200$

The z-score is,

$z = \frac{x - μ}{σ} = \frac{200 - 125}{7.9057} = 9.4868 z = 9.4868$

Therefore, the z-score is 9.4868.

Step 5: Finding the approximate probability

$d . P (x \leq 200) = P (x \leq 200) = P (z \leq 9.4868) = 1 P (x \leq 200) = 1$

Therefore, the approximate probability is 1.

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Suppose $χ$ is a normally distributed random variable with $μ = 50$ and $σ = 3$ . Find a value of the random variable, call it $χ_{0}$ , such that

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b) $P (χ > χ_{0}) = . 025$

c) $P (χ > χ_{0}) = . 95$

d) $P (41 \leq χ \leq χ_{0}) = . 8630$

e) 10% of the values of role="math" localid="1652160513072" $χ$ are less than role="math" localid="1652160519976" $χ_{0}$

f) 1% of the values of $χ$ are greater than $χ_{0}$

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