Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access)
Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access)
7th Edition
ISBN: 9781319019334
Author: David S. Moore, William I. Notz, Michael A. Fligner
Publisher: W. H. Freeman
Question
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Chapter 19, Problem 19.63SE

a.

To determine

Theprobability that exactly eight of the 12 visit an auction site in past month.

a.

Expert Solution
Check Mark

Answer to Problem 19.63SE

The approximate probability that exactly 8 out of 12 students visit an auction site is

0.1208.

Explanation of Solution

Given info:

In the age group 18 to 34, 50% of male internet users visit an auction site at least once a month. An interview done with 12 male internet users, out of those exactly eight visit the auction site in past month.

Calculation:

Convert the probability into decimal form,

p=50100=0.5

The formula to calculate probability that exactly 8 out of 12 students visit an auction site is,

P(X=k)=(nk)pk(1p)nk

Where n is number of units.

Simplify the above formula.

P(X=k)=(n!k!(nk)!)pk(1p)nk [(nk)=n!k!(nk)!]

Substitute 8 for k, 12 for n and 0.5 for p in above formula.

P(X=8)=(12!8!(128)!)(0.5)8(10.5)128=(12!8!4!)(0.00390625)(0.5)4=(9×10×11×121×2×3×4)(0.00390625)(0.0625)

Further simplify to find probability.

P(X=8)=495(2.4413×104)=0.1208

Thus, the approximate probability that exactly 8 out of 12 students visit an auction site is

0.1208.

b.

To determine

To find: The probability that at least 235 men in the sample visit online auction site at least once a month.

b.

Expert Solution
Check Mark

Answer to Problem 19.63SE

The probability that at least 235 men in the sample visit online auction site at least once a month is 0.9099.

Explanation of Solution

Given info:

In the age group 18 to 34, 50% of male internet users visit an auction site at least once a month. An interview done with 500 male internet users aged 18 to 34.

Calculation:

Convert the probability into decimal form,

p=50100=0.5

Use normal approximation equation to check that normal approximation is appropriate.

np10

Substitute 500 for n and 0.5 for p in above formula.

500×0.51025010

So, the normal approximation is correct.

Suppose, XB(n,p) and if n is large or p is 0.5, then XN(np,npq) .

The required probability is calculated as:

P(X>235)=P(Xnpnpq>235npnpq) (n=500,p=0.5,q=0.5)=P(X500×0.5500×0.5×0.5>235500×0.5500×0.5×0.5)=P(Z>1511.180)=P(Z>1.341)

Further solve the above expression.

P(X>235)=P(Z<1.341) (Bysymmetric)=0.9099 (Z-table)

Thus, the probability that at least 235 men in the sample visit online auction site at least once a month is 0.9099.

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Chapter 19 Solutions

Loose-leaf Version for The Basic Practice of Statistics 7e & LaunchPad (Twelve Month Access)

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