Question

Assume that the Gamecocks basketball team has a probability of 0.70 of winning a game against any opponent, and that the outcomes of its games are independent of each other. Suppose during the basketball season it will play a total of 30 games. Let X denote the total number of games that it will win during the season.

(a) Write down the formula for the probability mass function (pmf) (p(x)) of X. Is your pmf the binomial pmf?

(b) Use a computer (R will be good to use) or calculator to compute p(x) for each x = 0, 1, 2, . . . , 30. Then plot these probabilities with respect to x. Describe the shape of this PMF.

(c) Find Pr{X ≥ 25}.

(d) Find Pr{15 < X ≤ 20}.

(e) Find the mean µ of X.

(f) Find the variance σ 2 and standard deviation σ of X.

(g) Compute Pr{µ − 2σ ≤ X ≤ µ + 2σ}. Is this probability close to 0.95?

Answer 1

Part a)

Probability of success p = 0.70

Total number of trials n= 30

Since each trial is independent hence it is a Binomial Distribution and pmf is given by:

P[X=k]= k=0,1,2,3,...n

Part b

x   P[X=x]   P[X<=x]
0   2.05891E-16   2.05891E-16
1   1.44124E-14   1.46183E-14
2   4.87619E-13   5.02237E-13
3   1.06193E-11   1.11215E-11
4   1.67253E-10   1.78375E-10
5   2.02934E-09   2.20771E-09
6   1.97297E-08   2.19374E-08
7   1.57838E-07   1.79775E-07
8   1.05883E-06   1.2386E-06
9   6.03923E-06   7.27784E-06
10   2.95922E-05   3.68701E-05
11   0.000125543   0.000162413
12   0.000463811   0.000626224
13   0.001498467   0.002124691
14   0.004245656   0.006370346
15   0.010566965   0.016937311
16   0.023115236   0.040052548
17   0.044417513   0.08447006
18   0.074851734   0.159321795
19   0.110307819   0.269629614
20   0.141561701   0.411191315
21   0.157290779   0.568482094
22   0.150141198   0.718623292
23   0.121853726   0.840477018
24   0.08292823   0.923405248
25   0.046439809   0.969845057
26   0.020838376   0.990683433
27   0.007203389   0.997886822
28   0.001800847   0.999687669
29   0.000289792   0.999977461
30   2.25393E-05   1

Graph is approximately following normal distribution.

Part c)

Find Pr{X ≥ 25}= 0.07659475201

Part d)

Find Pr{15 < X ≤ 20}. =p[X=16]+ p[X=17] +p[X=18] +p[X=19] +p[X=20]

= 0.02311523619+0.04441751268+0.07485173433+0.11030781901+0.14156170106

= 0.39425400327