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Do you jog? The Gallup Poll once asked a random sample of1540adults, "Do you happen to jog?" Suppose that in fact 15%of all adults jog.

(a) What is the mean of the sampling distribution of p^? Justify your answer.

(b) Find the standard deviation of the sampling distribution of p^. Check that the 10%condition is met.

(c) Is the sampling distribution of p^approximately Normal? Justify your answer.

(d) Find the probability that between 13%and 17%of a random sample of 1540adults are joggers. Show your work.

Short Answer

Expert verified

a). The mean is0.15.

b). The standard deviation is0.0091.

c). Normal approximation might be appropriate.

d). The probability is0.9722.

Step by step solution

01

Part (a) Step 1: Given Information

Population proportion (p)=0.15,

Sample size (n)=1540.

02

Part (a) Step 2: Explanation

The sample distribution's mean can be computed as:

p^=p

=0.15

03

Part (b) Step 1: Given Information

The standard deviation of p^'s sampling distribution must be determined. Make sure that the 10% criteria is met.

04

Part (b) Step 2: Explanation

The sample proportion's standard deviation is calculated as:

p^=p(1-p)n

=0.15(1-0.15)1540

=0.0091

05

Part (c) Step 1: Given Information

Given in the question that, The Gallup Poll once asked a random sample of 1540adults.

We have to find is the sampling distribution of p close to Normal? Justify your response.

06

Part (c) Step 2: Explanation

Here,

np=1540(0.15)

=231>10

n(1-p)=1540(1-0.15)

=1309>10

As a result, normal approximation might be appropriate.

07

Part (d) Step 1: Given Information

We need to calculate the likelihood that between 13 and 17 percent of a random sample of1540 people are joggers. Display your work.

08

Part (d) Step 2: Explanation

The probability that between 13%and 17%of the sample of 1540adults are joggers is calculated as:

P(0.13p^0.17)=P0.13-0.150.0091Z0.17-0.150.0091

=P(-2.20Z2.20) =0.9722

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