Answer to Question #173574 in Statistics and Probability for ANJU JAYACHANDRAN

Question

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.

Answer 1

The distribution for the given data is-

Here, $\bar{x}=\dfrac{\sum X}{n}=\dfrac{340}{7}=48.57$

$\bar{Y}=\dfrac{\sum Y}{n}=\dfrac{1}{7}=0.1428$

Regression coefficient of y on x

$b_{yx}=\dfrac{n\sum XY-\sum X\sum Y}{n\sum X^2-(\sum X)^2}=\dfrac{7\times 46.8-340\times 1}{7\times (22400)-(340)^2}=\dfrac{-12.4}{41200}=-0.0003$

Regression equation of y on x -

$y-\bar{y}=b_{yx}(x-\bar{x})\\\Rightarrow y-0.1428=-0.0003(x-48.57)\\\Rightarrow y=-0.0003x+0.128$

The probability when runs scored-

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