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Mathematics 16 Online
OpenStudy (anonymous):

Find the probability of obtaining 25 samples with a mean height ranging from 60 to 63 inches from a normal population with a mean of 61 and a standard deviation of 5 inches?

OpenStudy (anonymous):

z = (X-Mean) / (SD/sqrt n) z1 = (60-61) / 5/sqrt 25 = - 1 z2 = (63-61) / (5/sqrt 25) = + 2 The area under the standard normal curve between these two z values indicates the required probability. Required probability = P(60 < Xbar < 63) = P(- 1 < z < 2) = 0.3413 (area corresponding to z1 being -ve lies on the left side of mean) + 0.4772 (area corresponding to z2 being +ve lies on the right side of mean) Is the answer 0.8185 ?

OpenStudy (welshfella):

I think so. But its been a long time.. I'll look something up.

OpenStudy (nincompoop):

yes

OpenStudy (nincompoop):

do you have your z-table values?

OpenStudy (anonymous):

the answer is correct? yay. Thankyouuu yeap. But I'm just confirming

OpenStudy (anonymous):

stat is really hard

OpenStudy (welshfella):

yes - i just looked up some old notes. Stats never came easy to me also.

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