Question

(a) The temperature (in °C) inside a reactor follows a probability distribution given by: (x-25) x > 25°C f(0) = 0 elsewhere e i. Verify that this is a valid probability density function, by way of relevant diagrams, calculations, and explanations. (Max 100 words) [5 marks] ii. Determine the median temperature inside the reactor, clearly justifying your answer. [5 marks] (b) The percent yield (Y) of product from a batch chemical process for 20 batches is given below: Y= {48.7 48.8 50.5 50.6 49.0 49.2 49.5 49.7 50.7 50.9 51.4 51.7 49.7 52.1 49.8 52.5 50.0 50.3 53.4 62.0} Draw a clearly labelled box plot for this data, showing how you calculate the various values. Comment on the distribution of the data presented with relevant justification. (Max 100 Words)

Answer 1

a)

reactor. » 1. F() 25 V a invous abo a) let X= temp (in °c) inside the Then pay of x : --x-25) f (2) e a) 25 elsewhere. Now the C.D.F of X be f (n) F(x) = Plxsx). (t-25) e dt. (-25) = Flu) 1 e x) 25 i) To check if its a valid pay. check the fonowing: - is fcal Continuous. a) - (-252 R is fin 2 is (25,00) -(x-25) = 3 ? x425 we on a on Continuous Thos,

& a in (25,–) f(x) > 0. -(-25) e -> &x in R So this property is abo satisfied f(x)dx=1. 25 20 -(1-25) e (x-257% > dx= 25 25 Eitte 479)+1 2) -25 - 2t e ti 7-700 e -25 2t No + -e a ♡ + 1 1. z This implies f(x) is a valid pay. = median of x. Son P(XEM) = 1 / 2 2 F ( = -ñ-25) - = -6 is such that

-(2-25) - 딧 늘 - । e-2 e 2 7 enfers = -- - 25 ) N 3. -In (P 2 In (2015) ८ ८) 27/28 in 2 + 25 25. 6931 (Ans)

b)

f. da (5) = (4) x(3) of x-1 Frequency) [d= fx-50.5 h f.d (2) 1= 50.5,1n =1 | (4) = (2) x(3) (3) .1.8 -1.8 (1) 48.7 3.24 1 1 1 48.8 -17 -17 289 2 49 1 -1.5 -1.5 225 3 49.2 1 -1.3 -1.3 1.69 4 49.5 -1 -1 1 5 49,7 -0.8 -0.8 0,64 6 49.7 -0.8 -0.8 0.64 7 1 1 1 1 1 1 49.8 -0.7 -0.7 0,49 8 50 .0.5 -0.5 0.25 9 50.3 -0.2 -0.2 0.04 10 50.5=A 1 0 0 0 11 50.6 0.1 0.1 0.01 12 50.7 0.2 0.2 0.04 13 50.9 0,4 0,4 0.16 14 1 1 1 1 1 1 514 0.9 0.9 0.81 15 51.7 1.2 1.2 1.44 16 52.1 1.6 1.6 2.56 17 52.5 2 2 4 18 1 1 53,4 29 29 8.41 19 62 1 11.5 11.5 132.25 20 ..- Pl =20 162.31 = f - d2ܐ 10.5 =f.dܐ -----


Σfd Mean x = A + n h 10.5 = = 50.5 + 1 20 = 50.5 +0.525 · 1 = 50.5 +0.525 = = 51.025 Median : in +1th M = value of observation 2 = value of 21 th 2 observation = value of 10.5th observation value of 10th observation + value of 11th observation 2 50.3 + 50.5 2 = 50.4 Mode : the frequency of observation 48.7 is maximum (1) : Z = 48.7


(Σf-d) 2 Σf. d. 2 Sample Variance s2 = h2 η - 1 (10.5)? 162.81 - 20 12 19 Ξ 162.81 - 5.5125 19 - 1 157.2975 1 19 = 8.2788 - 1 = 8.2788 (Σf d) 2 Σf d. Σ Sample Standard deviation S = h n - - 1 162.81 - (105)2 20 = 1 19


(98. d) Ef.d. 72 Sample Standard deviation S = .h n-1 162.81 - (10.5)2 20 1 19 Il 162.81 - 5.5125 19 1 157.2975 1 19 = 8.2788 1 = 2.8773 · 1 = 2.8773 S Co-efficient of Variation (Sample) = - · 100% x 2.8773 = 100% 51.025 = 5.64%

Descriptive Analysis:

The descriptive statistics for the sample are as folows

Sample size:20 Mean (*): 51.025 Median: 50.4 Mode: 49.7 Lowest value: 48.7 Highest value: 62 Range: 13.3 Interquartile range: 2.075 First quartile: 49.55 Third quartile: 51.625 Variance (s): 8.2788157894733 Standard deviation (s): 2.8772931358263 Quartile deviation: 1.0375 Mean absolute deviation (MAD): 1.695

The mean of the distribtuion is :51.025. With a s.d of 2.877. Hence the data is not very dispersed/scattered. The difference between the max and the min value in the data = range = 13.3. Half of the values in the data are below the median = 50.4 which is less than the mean. Also the mode is 49.7 is less than mean = The data is right skewed(positively) as Skewness: 3.168019.
And is leptokurtic with Kurtosis: 11.894859.

Box Plot:

Sample size: 20 Median: 50.4 Minimum: 48.7 Maximum: 62 First quartile: 49.55 Third quartile: 51.625 Interquartile Range: 2.075 Outlier: 62

162 161 1600 159. 1584 157 1561 1554 1544 1534 152 1514 1500 1491

The data has one outlier value = 62