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A large lecture theater has 270 seats, 24 of which can accomodate left-handed students. Suppose it is known that 14% of people are left-handed. One class held in the theater has 205 students.

(a) Let X be the number of left-handed students in the class. The mean and standard deviation of X are, respectively, and

(b) Let p hat be the proportion of left-handed students in the class. The mean and standard deviation of p hat are, respectively, and .

(c) What is the approximate probability that all left-handed students in the class get an appropriate desk?

I am assuming that this is supposed to be a hypergeometric probability that needs to be approximated to a normal distribution

mean is equation to 0.14*205 = 28.7

standard deviation is sqrt(205*.14*.86*((270-205)/(270-1)))=2.442140873

p hat = X/n

Not sure if that is even right.

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    $\begingroup$ I am wondering whether it should be binomial which needs to be approximated $\endgroup$ – Curious Nov 21 '14 at 4:33
  • $\begingroup$ @Curious I was wondering, I thought it was hypergeometric, but my standard deviation was wrong. So I am unsure $\endgroup$ – Bob Unger Nov 23 '14 at 3:17
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It's necessary to make certain assumptions.

If we assume that no right-handed person sits in a seat that can accommodate a left-hander, and we assume that people are only left- or right-handed (no ambidextrousness allowed), and we assume that the people in the class are effectively randomly selected with respect to handedness (e.g. no dependence due to being related or whatever) then the question reduces to:

"Given that 14% of people are left handed, out of 205 randomly selected people, what's the probability that no more than 24 will be left handed?"


Edit: explaining part c

(c) What is the approximate probability that all left-handed students in the class get an appropriate desk?

which I translated to:

"Given that 14% of people are left handed, out of 205 randomly selected people, what's the probability that no more than 24 will be left handed?"

Let $X$ be defined (as in your Q) as the number of left handed students in the class.

Under the assumptions we considered, $X$ is distributed $\text{binomial}(205,0.14)$, which is approximately $N(205 \times 0.14, 205 \times 0.14 \times (1-0.14))$ (i.e. $N(28.7,24.682)$).

$P(X\leq 24) = P(\frac{X-28.7}{\sqrt{24.682}} \leq \frac{24-28.7}{\sqrt{24.682}})\approx P(\frac{X-28.7}{\sqrt{24.682}} \leq -0.946) $

i. Ignoring continuity correction:

$P(X\leq 24) = P(\frac{X-28.7}{\sqrt{24.682}} \leq \frac{24-28.7}{\sqrt{24.682}})\approx P(\frac{X-28.7}{\sqrt{24.682}} \leq -0.946) \approx P(Z\leq -0.946) $

$\hspace{2.3cm}\approx 0.172$

ii. With continuity correction:

enter image description here

$P(X\leq 24) \approx P(\frac{X-28.7}{\sqrt{24.682}} \leq \frac{24+0.5-28.7}{\sqrt{24.682}})\approx P(Z\leq -0.845...) $

$\hspace{2.3cm}\approx 0.199$

iii. Exact binomial answer: $0.20081886...$

So it's around 20% chance.

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  • $\begingroup$ Thank you, and it should be approximated to binomial not hypergeometric. so the standard deviation should be 4.9861. for part b) the mean is 0.14 and the standard deviation is 4.9861/205. I didn't get c) right and i'm not sure why. I though it was the z score P(Z<(24-28.7)/4.9861) $\endgroup$ – Bob Unger Nov 23 '14 at 3:50
  • $\begingroup$ Yes, that's right - the point of me rewriting the question that was was to highlight that you needed the binomial (without just telling you how to do it). I can't do it right now but if I get a chance later I'll come back and explain (c) in detail. Did they use a continuity correction or not? $\endgroup$ – Glen_b Nov 23 '14 at 4:29
  • $\begingroup$ er, that first sentence should be "... that way was ...". I've now updated my answer with a fairly complete discussion of (c). I notice you have a typo at the end of your comment; the standard deviation is about 4.9681 not 4.9861 $\endgroup$ – Glen_b Nov 23 '14 at 7:24

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