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The heights of women in a certain area have a mean of 175cm and a standard deviation of 2.5cm. The heights of men in the same area have a mean of 177cm and a standard deviation of 2cm. Samples of 50 men and 40 women are taken and their heights are recorded. Find the probability that the mean height of men is more than 3cm greater than the mean height of women.

So far I have done this: $$W~N(175,2.5^2)$$ $$M~N(177,2^2)$$ $$M-W=X~N(2,\frac{10.25}{90})$$ $$P(X>3)=P(Z>\frac{3-2}{\sqrt{\frac{10.25}{90}}})$$ $$=P(Z>2.96)$$ $$=1-P(Z<2.96)$$

This does not lead to the right answer however. Can someone tell me where I've gone wrong.

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  • $\begingroup$ Where does the variance $\dfrac{10.25}{90}$ come from? $\endgroup$ – Did Apr 4 '15 at 20:15
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Distribution of the difference of normal sample means.

This is getting to be too scattered a trail between what you asked to have checked and various suggestions. You have given it a good try, but have not properly taken averaging into account. Here is an outline to follow. You will need to fill in some gaps, including giving reasons to justify some of the steps.

$$W \sim N(175, 2.5^2), \;\;\mathrm{so}\;\; \bar W \sim N(175, 2.5^2/40).$$ $$M \sim N(177. 2^2), \;\;\mathrm{so}\;\;\bar M \sim N(177, 2^2/50).$$

Under appropriate assumptions (which you should explain): $$D = \bar M - \bar W \sim N(177 - 175, \frac{2^2}{50} + \frac{2.5^2}{40}) = N(2, 0.225).$$

You seek $P\{D > 3\} = P\{Z > (3-2)/\sqrt{0.225}\},$ where $Z$ is standard normal. Then get the answer from printed normal tables. (Hint: It is somewhere between 0.017 and 0.019.)

Notes: (1) @Did was correct to question a variance with 90 in the denominator. (2) To get a tilde $\sim$ inside a TeX expression, you have to use \sim . (3) In real life, it would be quite unusual for the population with the larger mean to have the smaller variance. Also, a standard deviation of 2 would imply that almost all women have heights in a range of 175 $\pm$ 6 cm, which is not realistic.

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  • $\begingroup$ If this is enough help, please click to 'Accept' and/or up vote it, and get this out of our 'no satisfactory answer' queue. Otherwise, ask a question and I'll check back soon. $\endgroup$ – BruceET Apr 4 '15 at 22:45
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Hint: what is the distribution of the mean of both samples?

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  • $\begingroup$ Normal? But how does that help? $\endgroup$ – Andrew Apr 4 '15 at 15:14
  • $\begingroup$ normal with which mean and std? $\endgroup$ – mookid Apr 4 '15 at 15:19
  • $\begingroup$ Mean$=2$ and Standard Deviation $=\sqrt{\frac{10.25}{90}}$ $\endgroup$ – Andrew Apr 4 '15 at 15:32
  • $\begingroup$ so what is the distribution fo the difference? $\endgroup$ – mookid Apr 4 '15 at 15:45
  • $\begingroup$ $$X~N(2,\frac{10.25}{90})$$ $\endgroup$ – Andrew Apr 4 '15 at 15:47

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