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OpenStudy (anonymous):

In a survey, 600 mothers and fathers were asked about the importance of sports for boys and girls. Of the parents interviewed, 70% said the genders are equal and should have equal opportunities to participate in sports. Using the normal approximation, what's the probability a randomly drawn sample of parents of size 600 will have a sample proportion between 67% and 73%? Draw a sketch of the probability curve, shade the area representing the probability you're finding, and label the z-scores that represent the upper and lower bounds of the probability you're finding.

OpenStudy (anonymous):

the mean is .7 and the sd=0.0187 N(.7,0.0187) I just don't know what to do from here, also can't use the continuity correction.

OpenStudy (faiqraees):

so are you aware with normal distribution?

OpenStudy (anonymous):

i know what the normal distribution is if thats what your asking? @faiqfaees

OpenStudy (faiqraees):

okay very good. |dw:1455459145428:dw| can you calculate the shaded area

OpenStudy (anonymous):

you would need to find the zscores of each right? @faiqraees

OpenStudy (faiqraees):

yes

OpenStudy (faiqraees):

oh i am sorry 0.63 will lie on the other side (negative x axis)

OpenStudy (anonymous):

I got -3.743315508 for .63 and 1.604278075 for .73 is that right?

OpenStudy (anonymous):

@faiqraees

OpenStudy (faiqraees):

well I actually dont have access to the areas to the values of z right now but if you send me the method you applied I can verify it for you.

OpenStudy (anonymous):

i just did x-mean/sd for both @faiqraees

OpenStudy (faiqraees):

then it should be correct

OpenStudy (anonymous):

okay so are those the zscore values that represent the lower and upper bounds of the probabiliity? @faiqraees

OpenStudy (faiqraees):

now use the identity total probability = (first prob. +second prob.) - 1

OpenStudy (faiqraees):

that way you will have the answer

OpenStudy (anonymous):

okay so -3.743315508 + 1.604278075 -1 = -3.139037433 @faiqraees so is that the probability of drawing between 67% and 73%?

OpenStudy (faiqraees):

oh no no after taking those z values you have to calculate their areas using the normal distribution function chart.

OpenStudy (anonymous):

okay so what do I do with the -3.139037433 or did i not need to find that @faiqraees

OpenStudy (faiqraees):

you dont need that just find the normal distribution values for -3.743315508 and 1.604278075

OpenStudy (anonymous):

okay so just find their z-scores then? @faiqraees

OpenStudy (faiqraees):

no find the area under the curve if you integrate the function with negative infinity and z value

OpenStudy (faiqraees):

do you have a normal distribution function chart or something

OpenStudy (anonymous):

no i dont but i can probably find one on the internet @faiqraees

OpenStudy (anonymous):

i dont know how to find one for -3.74 but for 1.604278075 i got .4452 but i dont know if thats right? @faiqraees

OpenStudy (faiqraees):

for 1.6 one the answer is 0.9456 and i cant also find the values for -3.74. Its above the range of my chart. Try googling its value

OpenStudy (anonymous):

for -3.7 it says .00011 but that doesnt seem right @faiqraees

OpenStudy (faiqraees):

there are many kind of normal distribution are you sure you're checking the right one?

OpenStudy (anonymous):

It just said standard normal distribution table: Table Values Represent AREA to the LEFT of the Z score. I don't know haha @faiqraees

OpenStudy (faiqraees):

how about checking the value of 1.604278075 through it. If it shows the value you're concerned with then thats the correct answer

OpenStudy (anonymous):

for 1.6 is shows .94520 @faiqraees

OpenStudy (faiqraees):

ok wait let me work out the value using integration

OpenStudy (faiqraees):

for -3.74 just do 1- 0 .00011

OpenStudy (anonymous):

okay so we have .9456 and 0.99989 so would we add them then subtract 1? @faiqraees

OpenStudy (faiqraees):

yes to find the answer

OpenStudy (anonymous):

okay thank you so much

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