Answer:
a
[tex]P( \^ p > 0.775 ) = 0.12798[/tex]
b
[tex]P( 0.6718 < p < 0.775 ) =0.87183[/tex]
Step-by-step explanation:
From the question we are told that
The population proportion is [tex]p = 0.75[/tex]
Considering question a
The sample size is [tex]n = 387[/tex]
Generally the standard deviation of this sampling distribution is
[tex]\sigma = \sqrt{ \frac{p(1 - p)}{ n } }[/tex]
=> [tex]\sigma = \sqrt{ \frac{0.75(1 - 0.75)}{ 387 } }[/tex]
=> [tex]\sigma = 0.022[/tex]
The sample proportion of cans that are recycled is
[tex]\^ p = \frac{ 300}{387 }[/tex]
=> [tex]\^ p = 0.775[/tex]
Generally the probability that 300 or more will be recycled is mathematically represented as
[tex]P( \^ p > 0.775 ) = P( \frac{\^ p - p }{ \sigma } > \frac{0.775 - 0.75 }{ 0.022} )[/tex]
[tex]\frac{\^ p - p }{\sigma } = Z (The \ standardized \ value\ of \ \^ p )[/tex]
[tex]P( \^ p > 0.775 ) = P( Z > 1.136 )[/tex]
From the z table the area under the normal curve to the left corresponding to 1.591 is
[tex]P( Z > 1.136) = 0.12798[/tex]
=> [tex]P( \^ p > 0.775 ) = 0.12798[/tex]
Considering question b
Generally the lower limit of sample proportion of cans that are recycled is
[tex]\^ p_1 = \frac{ 260 }{387 }[/tex]
=> [tex]\^ p_1 = 0.6718[/tex]
Generally the upper limit of sample proportion of cans that are recycled is
[tex]\^ p_2 = \frac{ 300}{387 }[/tex]
=> [tex]\^ p_2 = 0.775[/tex]
Generally probability that between 260 and 300 will be recycled is mathematically represented as
[tex]P( 0.6718 < p < 0.775 ) = P( \frac{0.6718 - 0.75 }{ 0.022}< \frac{\^ p - p }{ \sigma } <\frac{0.775 - 0.75 }{ 0.022} )[/tex]
=> [tex]P( 0.6718 < p < 0.775 ) = P( -3.55 < Z < 1.136 )[/tex]
=> [tex]P( 0.6718 < p < 0.775 ) = P(Z < 1.136 ) - P( Z < -3.55 )[/tex]
From the z table the area under the normal curve to the left corresponding to 1.136 and -3.55 is
[tex]P( Z < -3.55 ) = 0.00019262[/tex]
and
[tex]P(Z < 1.136 ) = 0.87202[/tex]
So
[tex]P( 0.6718 < p < 0.775 ) = 0.87202- 0.00019262[/tex]
=> [tex]P( 0.6718 < p < 0.775 ) =0.87183[/tex]
