Question

Assume that 25% of all households in Santa Clara county have a burglar alarm (population characteristic). A random sample of 200 Santa Clara county households is taken. What is the probability that between 20% (0.2) to 30% (0.3) of the households in the sample have a burglar alarm

Answer 1

The probability is
`P( 0.2 < p < 0.3) = 0.89751`

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
`\mu_(x) = p = 0.25`

Generally the standard deviation is mathematically represented as

`\sigma = \sqrt{ ( p (1- p ) )/(n) }`

=>
`\sigma = \sqrt{ ( 0.25(1- 0.25 ) )/(200) }`

=>
`\sigma = 0.03062`

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

`P( 0.2 < p < 0.3) = P(( 0.2 - 0.25)/( 0.03062 ) < (p- \mu_(x))/(\sigma ) < ( 0.2 - 0.25)/( 0.03062 ) )`

`(p -\mu)/(\sigma )  =  Z (The  \ standardized \  value\  of  \  p )`

=>
`P( 0.2 < p < 0.3) = P(-1.6329 < Z <1.6329 )`

=>
`P( 0.2 < p < 0.3) = P( Z< 1.6329) - P( Z <- 1.6329 )`

From the z table the area under the normal curve to the left corresponding to 1.6329 and -1.6329 is

`P( Z< 1.6329) = 0.94875`

and

`P( Z <- 1.6329 ) = 0.051245`

So

`P( 0.2 < p < 0.3) = 0.94875 - 0.051245`

=>
`P( 0.2 < p < 0.3) = 0.89751`