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Question # 4 (A) The life-time of a perishable product specified in months is modeled by a random variable (RV), x with a pdf given by: p(x) = k × (9-x) for 0 otherwise x S9 Here, k is a positive constant. Find: (a) Value of k; and, (b) expected value of the life-time is less than 4 months. life-time; and, (c) probability that the lAnswer-hints: (a) k -2/81; ( p(0 SS4) 56/81 A set of computer product manufactured is tested. Suppose x denotes the number of units passing the test. (a) Find the variance of x and p(x) depicts the associated probability as listed below (Data set I; and, plot the CD (B.1) Data set I x in 1000s012 3 pex 1/8 3/8 3/8 18 Answer-hints: V(x) 3/41 (B.2) In Question 4(B.1) above, suppose the data set changes as follows s in 1000s | O L 2 plex) Data set I 0.25 0.25 0.25 0.25 Data set III x in 1000s 0 2 3 0.15 0.25 0.35 0.25 Determine variances of these new data sets and plot the CDF in each case. Hence, explain the salient differences between the statistics of Data sets I, II and III in terms of the associated kurtosis and skew.

Answer 1

(B.1)

(a)

E[X] = $\sum x_i * p(x_i)$

= 0 * (1/8) + 1 * (3/8) + 2 * (3/8) + 3 * (1/8) = 1.5

E[X²] = $\sum x_i^2 * p(x_i)$

= 0² * (1/8) + 1² * (3/8) + 2² * (3/8) + 3² * (1/8) = 3

Variance = E[X²] - E[X]² = 3 - 1.5²  = 0.75 = 3/4

CDF's are plotted in below image.

(B.2)

Data Set II

E[X] = $\sum x_i * p(x_i)$

= 0 * 0.25 + 1 * 0.25 + 2 * 0.25 + 3 * 0.25 = 1.5

E[X²] = $\sum x_i^2 * p(x_i)$

= 0² * 0.25 + 1² * 0.25 + 2² * 0.25 + 3² * 0.25 = 3.5

Variance = E[X²] - E[X]² = 3.5 - 1.5²  = 1.25 = 5/4

Data Set II|

E[X] = $\sum x_i * p(x_i)$

= 0 * 0.15 + 1 * 0.25 + 2 * 0.35 + 3 * 0.25 = 1.7

E[X²] = $\sum x_i^2 * p(x_i)$

= 0² * 0.15 + 1² * 0.25 + 2² * 0.35 + 3² * 0.25 = 3.9

Variance = E[X²] - E[X]² = 3.9 - 1.7²  = 1.01

CDF's are plotted in below image.

skewness: g₁ = m₃ / m₂^3/2

kurtosis: a₄ = m₄ / m₂²

where m2, m3 and m4 are second, third and fourth central moments of the distribution.

m2 = variance.

Dataset I

m3 = E[(X - $\mu$)³] = $\sum (x_i - \mu)^3 * p(x_i)$

= (0 - 1.5)³ * (1/8) + (1 - 1.5)³ * (3/8) + (2 - 1.5)³ * (3/8) + (3 - 1.5)³ * (1/8) = 0

m4 = E[(X - $\mu$)⁴] = $\sum (x_i - \mu)^4 * p(x_i)$

= (0 - 1.5)⁴ * (1/8) + (1 - 1.5)⁴ * (3/8) + (2 - 1.5)⁴ * (3/8) + (3 - 1.5)⁴ * (1/8) = 1.3125

skewness: g₁ = m₃ / m₂^3/2 = 0 / 0.75^3/2 = 0

kurtosis: a₄ = m₄ / m₂² = 1.3125 / 0.75² = 2.33

Dataset II

m3 = E[(X - $\mu$)³] = $\sum (x_i - \mu)^3 * p(x_i)$

= (0 - 1.5)³ * 0.25 + (1 - 1.5)³ * 0.25 + (2 - 1.5)³ * 0.25 + (3 - 1.5)³ * 0.25 = 0

m4 = E[(X - $\mu$)⁴] = $\sum (x_i - \mu)^4 * p(x_i)$

= (0 - 1.5)⁴ * 0.25 + (1 - 1.5)⁴ * 0.25 + (2 - 1.5)⁴ * 0.25 + (3 - 1.5)⁴ * 0.25 = 2.5625

skewness: g₁ = m₃ / m₂^3/2 = 0 / 1.25^3/2 = 0

kurtosis: a₄ = m₄ / m₂² = 2.5625 / 1.25² = 1.64

Dataset III

m3 = E[(X - $\mu$)³] = $\sum (x_i - \mu)^3 * p(x_i)$

= (0 - 1.5)³ * 0.15 + (1 - 1.5)³ * 0.25 + (2 - 1.5)³ * 0.35 + (3 - 1.5)³ * 0.25 = 0.35

m4 = E[(X - $\mu$)⁴] = $\sum (x_i - \mu)^4 * p(x_i)$

= (0 - 1.5)⁴ * 0.15 + (1 - 1.5)⁴ * 0.25 + (2 - 1.5)⁴ * 0.35 + (3 - 1.5)⁴ * 0.25 = 2.0625

skewness: g₁ = m₃ / m₂^3/2 = 0.35 / 1.01^3/2 = 0.3448

kurtosis: a₄ = m₄ / m₂² = 2.0625 / 1.01² = 2.02

Since the skewness of dataset I and II is 0, the distribution of dataset I and II is symmetric. The the distribution of dataset III is moderately skewed.

Since the kurtosis is less than 3 for all the datasets, the distribution of all datasets is platykurtic. Compared to a normal distribution, the tails of the distribution will be shorter and thinner, and often its central peak is lower and broader.