Respuesta :
Let p be
the population proportion.
We have p=0.60, n=200 and we are asked to find
P(^p<0.58).
The thumb of the rule is since n*p = 200*0.60
and n*(1-p)= 200*(1-0.60) = 80 are both at least greater than 5, then n is
considered to be large and hence the sampling distribution of sample
proportion-^p will follow the z standard normal distribution. Hence this
sampling distribution will have the mean of all sample proportions- U^p = p =
0.60 and the standard deviation of all sample proportions- δ^p = √[p*(1-p)/n] =
√[0.60*(1-0.60)/200] = √0.0012.
So, the probability that the sample proportion
is less than 0.58
= P(^p<0.58)
= P{[(^p-U^p)/√[p*(1-p)/n]<[(0.58-0.60)/√0...
= P(z<-0.58)
= P(z<0) - P(-0.58<z<0)
= 0.5 - 0.2190
= 0.281
So, there is 0.281 or 28.1% probability that the
sample proportion is less than 0.58.
There is 0.281 or 28.1% probability that the sample proportion is less than 0.58.
Let p be the population proportion.
We have given that,
p=0.60,
n=200
and we are asked to find[tex]P(p < 0.58).[/tex]
What is the thumb rule?
The thumb of the rule is n*p.
by using the thumb rule n*p = 200*0.60
and n*(1-p)
= 200*(1-0.60)
= 80
That are both at least greater than 5, then n is considered to be large and hence.
The sampling distribution of sample proportion power p will follow the z standard normal distribution.
Hence this sampling distribution will have the mean of all sample proportions-
[tex]U^p = p = 0.60[/tex]
and the standard deviation of all sample proportions
[tex]\delta^p = \sqrt{[p\times(1-p)/n]}\\\\ = \sqrt{[0.60\times (1-0.60)/200]}\\\\\\= \sqrt {0.0012.}[/tex]
So, the probability that the sample proportion is less than 0.58
= [tex]=P(p < 0.58)[/tex]
[tex]= P{[(^p-U^p)/ \sqrt{[p*(1-p)/n]} < [(0.58-0.60)/\sqrt0...[/tex]
[tex]= P(z < -0.58)[/tex]
[tex]= P(z < 0) - P(-0.58 < z < 0)[/tex]
[tex]= 0.5 - 0.2190[/tex]
[tex]= 0.281[/tex]
Therefore there is 0.281 or 28.1% probability that the sample proportion is less than 0.58.
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