Introduction to Statistics PS5 PDF

Preview

Summary

Download Introduction to Statistics PS5 PDF

Description

Introduction to Statistics (S520) Problem Set 5 (Due date: 28th September)

1.

a) To find Median(q2): Area under the graph [0,q2] = 0.5 1*0.3 + (q2 – 1)*0.7 = 0.5 q2 – 1 = 0.2/0.7 q2 = 2/7 + 1 q2 = 9/7 ∞

b) E(X) =

∫

1

xf (x ) dx =

−∞

∫ 0.3 xdx

2

+

0

= 0.3 (12 – 02) + 0.7 (22 – 12) = 0.3 + 2.1 = 2.4

∫ 0.7 xdx 1

2.

0.5 * (q2 – 1) * 2 * (q2 – 1) = 0.5 (q2 – 1)2 = 0.5 q2 = 0.707 + 1 (root cannot be negative since q2>1 from graph) q2 (population median) = 1.707 (upto 3 decimal places)

0.5 * (q1 – 1) * 2 * (q1 – 1) = 0.25 (q1 – 1)2 = 0.25 q1 = 0.5 + 1 (root cannot be negative since q1>1 from graph) q1 = 1.5 0.5 * (q3 – 1) * 2 * (q3 – 1) = 0.75 (q3 – 1)2 = 0.75 q3 = 0.866 + 1 (root cannot be negative since q2>1 from graph) q3 = 1.866 (upto 3 decimal places) Therefore, Iqr(X) = q3 – q1 = 1.866 – 1.5 = 0.366

3.

a) TRUE. The q1 and q3 are equidistant from the median for a symmetric random variable and thus the median will be equal to their average. b) FALSE. For a symmetric random variable, the inter quartile range is always greater than the standard deviation, but in other cases, the standard deviation can exceed the inter quartile range. c) FALSE. In case the distribution has extreme outliers, the expected value of X can lie outside the range between the first and the third quartile of X. d) TRUE. If the standard deviation is 0, it means that the random variable assumes the same value for all values of x and thus has the same q1 and q3. Therefore, the inter quartile range (q3 – q1) equals 0. e) FALSE. In case of a symmetric random variable, the mean is always equal to the median but the converse is not always true.

4.

qnorm(0.025,mean = 85,sd = 5) = 75.20018 Therefore, the manager will have to reduce the price for weekly demand below 75.20018 lbs.

5.

qnorm(0.999999,mean = 100,sd = 20) = 195.0685 Therefore Lulu’s CCQ score = 195.0685

R output for 4 and 5:

6.

to find qx, Y = X2 P(Y...