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According to the U.S. census, the proportion of adults in a certain county who owned their own home was 0.71. An SRS of 100adults in a certain section of the county found that65 owned their home. Which one
of the following represents the approximate probability of obtaining a sample of 100adults in which fewer than 65own their home, assuming that this section of the county has the same overall proportion of adults who own their home as does the entire county?

(a) 10065(0.71)65(0.29)35

(b) 10065(0.29)65(0.71)35
(c) Pz<0.65-0.71(0.71)(0.29)100

(d) Pz<0.65-0.71(0.65)(0.35)100
(e) Pz<0.65-0.71(0.71)(0.29)100

Short Answer

Expert verified

The correct answer is option (c)Pz<0.65-0.71(0.71)(0.29)100.

Step by step solution

01

Given information

The proportion of adults who owned their own home was 0.71.An SRS of 100adults of the county found that 65owned their home.

02

Explanation

Let, sample sizen=100
Number of successesx=65
Population proportion p=0.71
Requirements for a normal approximation of the binomial distribution: np≥10andnq≥10.
np=100(0.71)=71≥10nq=n(1-p)=100(1-0.71)=29≥10

The proportion is:
p^=xn=65100=0.65
Then the mean is:
μp^=p=0.71

03

Calculation

The standard deviation is:
σp^=p(1-p)n=0.71(1-0.71)100=0.71(0.29)100

The z-score is:

z=x-μσ=0.65-0.710.71(0.29)100

To determine the probability that the sample proportion is less than 0.65.

P(p^<0.65)=PZ<0.65-0.710.71(0.29)100

Hence, option (c) Pz<0.65-0.71(0.71)(0.29)100is correct.

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