Answer:
The probability is [tex]P( 0.2 < p < 0.3) = 0.89751[/tex]
Step-by-step explanation:
From the question we are told that
The population proportion is p = 0.25
The sample size is n = 200
Generally given that the sample size is large the mean of this sampling distribution is mathematically represented as [tex]\mu_{x} = p = 0.25[/tex]
Generally the standard deviation is mathematically represented as
[tex]\sigma = \sqrt{ \frac{ p (1- p ) }{n} }[/tex]
=> [tex]\sigma = \sqrt{ \frac{ 0.25(1- 0.25 ) }{200} }[/tex]
=> [tex]\sigma = 0.03062[/tex]
Generally the probability that between 20% (0.2) to 30% (0.3) of the households in the sample have a burglar alarm is mathematically represented as
[tex]P( 0.2 < p < 0.3) = P(\frac{ 0.2 - 0.25}{ 0.03062 } < \frac{p- \mu_{x}}{\sigma } < \frac{ 0.2 - 0.25}{ 0.03062 } )[/tex]
[tex]\frac{p -\mu}{\sigma } = Z (The \ standardized \ value\ of \ p )[/tex]
=> [tex]P( 0.2 < p < 0.3) = P(-1.6329 < Z <1.6329 )[/tex]
=> [tex]P( 0.2 < p < 0.3) = P( Z< 1.6329) - P( Z <- 1.6329 )[/tex]
From the z table the area under the normal curve to the left corresponding to 1.6329 and -1.6329 is
[tex]P( Z< 1.6329) = 0.94875[/tex]
and
[tex]P( Z <- 1.6329 ) = 0.051245[/tex]
So
[tex]P( 0.2 < p < 0.3) = 0.94875 - 0.051245[/tex]
=> [tex]P( 0.2 < p < 0.3) = 0.89751[/tex]
