Statistical Probabilities and Distributions

Odd Solutions for Stat. Prob./Dist. Homework 9

1. Separately calculate using the binomial formula the probabilities of getting 0, 1, 2, 3, or 4 left-handed students in a class of 25, given a probability of 0.1. Compare your results with those obtained by doing binompdf(25,.1) (DISTR 0) on your TI-83 calculator.

    

   x	P(x)	25Cxpxqn-x
   0	0.07179	1 • 0.10 • 0.925
   1	0.19942	25 • 0.11 • 0.924
   2	0.26589	300 • 0.12 • 0.923
   3	0.22650	2300 • 0.13 • 0.922
   4	0.13842	12650 • 0.14 • 0.921
   5	0.06459	53130 • 0.15 • 0.920

2.  

3. Check the assumptions carefully and see if we are justified in using the binomial (and not the hypergeometric) distribution for the problems above.

     Sample likely less than 10% of population.

5. Calculate the probability described in the text for winning the lottery by matching 5 of the 6 selected numbers from 54.

     P(x=5)=[6!/(1!5!)] • [48!/(47!1!)] ÷ [54!/(48!6!)]=1.12E-5

7. Use the normal approximation for the binomial to calculate the probability of getting 10 heads in 20 attempts from a fair coin (ignore the magic number test). Be sure to use the continuity correct and calculate the area under the probability density curve from 9.5 to 10.5. Compare this carefully with the results from the binomial formula.

     mean=n • p=10           s.d.=sqrt(n • p • q)=2.236
   

   x	x-0.5	x+0.5	z(x-0.5)	z(x+0.5)	P(x+/-0.5)	20Cx0.520
   10	9.5	10.5	-0.224	0.224	0.177	0.176
   11	10.5	11.5	0.224	0.671	0.160	0.160
   12	11.5	12.5	0.671	1.118	0.119	0.120
   13	12.5	13.5	1.118	1.565	0.0730	0.0739
   14	13.5	14.5	1.565	2.013	0.0367	0.0370
   15	14.5	15.5	2.013	2.460	0.0151	0.0148
   16	15.5	16.5	2.460	2.907	0.00512	0.00462
   17	16.5	17.5	2.907	3.354	0.00143	0.00109
   18	17.5	18.5	3.354	3.801	3.26E-4	1.81E-4
   19	18.5	19.5	3.801	4.249	6.13E-5	1.91E-5
   20	19.5	20.5	4.249	4.696	9.42E-6	9.54E-7

   Only a few percent error until more than 2 s.d. from mean.

9. Use the normal approximation for the binomial to calculate the probability of getting 13 heads in 20 attempts (ignore the magic number test). Compare this carefully with the results from the binomial formula. Is this the same as the probability of getting 7 heads?

     See above. Yes.

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