Question #173574

6(b) Based on the previous data, the probabilities of a batsman making various scores in

One Day Internationals are given below: (5)

Runs 10 20 30 50 60 70 100

Probability 0.01 0.20 0.15 0.30 0.12 0.2 0.02

Simulate the runs scored by the batsman in the next five One Day Internationals using

the following 25, 39, 65, 76, 12.


1
Expert's answer
2021-05-11T08:50:16-0400

The distribution for the given data is-





Here, xˉ=Xn=3407=48.57\bar{x}=\dfrac{\sum X}{n}=\dfrac{340}{7}=48.57


Yˉ=Yn=17=0.1428\bar{Y}=\dfrac{\sum Y}{n}=\dfrac{1}{7}=0.1428


Regression coefficient of y on x


byx=nXYXYnX2(X)2=7×46.8340×17×(22400)(340)2=12.441200=0.0003b_{yx}=\dfrac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\dfrac{7\times 46.8-340\times 1}{7\times (22400)-(340)^2}=\dfrac{-12.4}{41200}=-0.0003


Regression equation of y on x -


yyˉ=byx(xxˉ)y0.1428=0.0003(x48.57)y=0.0003x+0.128y-\bar{y}=b_{yx}(x-\bar{x})\\\Rightarrow y-0.1428=-0.0003(x-48.57)\\\Rightarrow y=-0.0003x+0.128



The probability when runs scored-


25 is y=0.0003(25)+0.128=0.120525 is y=0.0003(39)+0.128=0.116325 is y=0.0003(65)+0.128=0.108525 is y=0.0003(76)+0.128=0.105225 is y=0.0003(12)+0.128=0.124425 \text{ is } y=-0.0003(25)+0.128=0.1205 \\[9pt] 25 \text{ is } y=-0.0003(39)+0.128=0.1163 \\[9pt] 25 \text{ is } y=-0.0003(65)+0.128=0.1085 \\[9pt] 25 \text{ is } y=-0.0003(76)+0.128=0.1052 \\[9pt] 25 \text{ is } y=-0.0003(12)+0.128=0.1244


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