MS-8 1.
Calc Calcul ulat ate e the the mean mean,, medi median an and and mod mode e from from the the foll follow owin ing g data data rel relat atin ing g to pro produ duct ctio ion n of a ste steel el mill for 60 days Production (in tons per day) 21-22 23-24 25-26 27-28 29-30 Number of days 7 13 22 10 8
Ans:
Mean: Mean: Let a = 25.5, h = 2 Clas lass in interval l( l(C.I) 21 - 22 23 – 24 25 – 26 27 – 28 29 – 30
Frequency (f (f i) 7 13 22 10 8 Σ f i = 60
xi 21.5 23.5 25.5 = a 27.5 29.5
ui= (x – 25.5)/2 -2 -1 0 1 2
fiu i -14 -13 0 10 16 Σ fiui = -1
By step-deviation method, Mean (x) ͞ = a + [Σ f iui/ Σ f i] x h = 25.5 + (-1/60)2 = 25.5 – 1/30 = 25.5 – 0.0333 = 25.4667 Mean production of steal is 25.4667 tons per day Median: Median: Making class intervals continuous by subtracting 0.5 from each lower limit and adding 0.5 to the upper class limit. Required cumulative frequency is: Production 21 - 22 23 – 24 25 – 26 27 – 28 29 – 30
C.I 20.5 22.5 24.5 26.5 28.5
Number of days
– 22.5 – 24.5 – 26.5 – 28.5 – 30.5
7 13 22 10 8 N = 60
Cumulative frequency 7 20 42 52 60
For median, N/2 = 30 which lies in the class 24.5 – 26.5 Here l = 24.5, f = 22, cf = 20, h = 2 Median = l + [(N/2 – cf)/f]xh = 24.5 + [(30 – 20)/22] x 2 = 24.5 + 20/22 = 24.5 + 0.90909091 = 25.41 Mode: Mode: Since the class 24.5 – 26.5 has the maximum observations, i.e., 22 f 1 = 22, f 0 = 13, f 2 = 10, l = 24.5 Mode = l + [(f 1 - f 0)/(2f 1 - f 0 - f 2)]h = 24.5 + [(22 – 13)/(2 x 22 – 13 – 10)] x 2 = 24.5 + (9/21) x 2 = 24.5 + 0.8571 = 25.357 -----x-----x-----x-----
2.
A rest restau aura rant nt is is expe experi rien ence ced d disco discont nten entm tmen entt amon among g its cus custo tome mers rs.. It anal analys yses es tha thatt there there are are three factors responsible viz. food quality, service quality and interior décor. By conducting an analysis, it assesses the probabilities of discontentment with the three factors as 0.40, 0.35 and 0.25 0.25 respec respectiv tively ely.. By conduc conductin ting g a survey survey among among the custom customers ers,, it also also evalua evaluated ted the
1
probabilities of a customer going away discontented on account of these factors as 0.6, 0.8 and 0.5, respectively. With this information, the restaurant wants to know that, if a customer is discontented, what are the probabilities that it is so due to food, service or interior décor? Ans:
Let
P( E1) : Discontentment with the food quality = 0.40 P(E2) : Discontentment with the service quality = 0.35 P(E3) : Discontentment with with the interior décor = 0.25 Let A : Customer going away discontented And P(A/E1) = 0.6, P(A/E2) = 0.8, P(A/E3) = 0.5 P(E1/A) = ?, P(E2/A) = ?, P(E3/C) = ? Using Baye’s theorem, P(E1/A) =
=
=
= 0.24/0.645 = 0.372
P(E2/A) =
=
=
= 0.28/0.645 = 0.434
P(E3/A) =
=
=
= 0.125/0.645 = 0.77
Discontentment due to food is 0.372, due to service quality 0.434 and due to interior décor 0.77 -----x-----x-----x-----
3.
The The mont monthly hly inco income mes s of a gro group up of of 10,0 10,000 00 per perso sons ns wer were e foun found d to be be norm normal ally ly dis distr tribu ibute ted d with with mean equal to 15,000 and standard deviation equal to 1000. What is the lowest income among the richest 250 persons?
Ans: Ans:
Here Here aver averag age e mont monthl hly y inco income me µ = 15,0 15,000 00 Standard deviation σ = 1000 We have to find lowest income among the richest 250 persons If we take x as a random variable depicting the lowest income, we have to find Pr(x < 250)
2
We have to find Pr(x > 9750) i.e. Pr (
= Pr (
< <
) = Pr ( <
)
) = Pr (Z < -5.25) = 0.5 – Pr (Z < -5.25)
= 0.5 – Pr ( -5.25 < Z < 0) = 0.5 – Pr (0 < Z < 5.25) = 0.5 – 5.25 = 4.75 = 18080 (From normal table) lowest income among the richest 250 persons =
18080
4.
Write short notes on the following: (a) Test of goodness of fit (b) Critical Region of a test (c) Exponential sm smoothing me method
Ans:
(a) Test of goodness of fit: fit: Population distribution is Normal, Poisson, and Uniform. This will base on evidence produced by a sample. This procedure is developed to test how close the fit
3
between the observed data and the distribution assumed is. These tests are based on the chisquare statistic. (b) Critical Region of a test: test: The sample mean would be used to draw conclusions about the population mean and so the test statistic is R. We shall be in a position to reject Ho only if the sample evidence is strongly against it i.e. if the observed value of x is much larger than 20. The critical region will therefore be of the form: x? c, where c is a real number much larger than 20. The actual value of c would depend on the significance level used.
This has been shown as the shaded region in Figure V above, where the distribution of has been been shown shown as a normal normal curve. curve. This This is valid valid under under two conditio conditionsns-(1) (1) if the populatio population n distribution is normal, then the distribution of z is also normal, or (2) if the 'sample size is large, then again, the central limit theorem assures us that the distribution of x can be approximated by a normal distribution. Therefore, if either of these conditions is valid (and in this case the second condition is certainly valid as n = 100), then
Now that we have identified the critical region, we can compare the observed value of x and see if it belongs to critical region. The observed value of x is 20.5-which lies in the critical region and so we can conclude that the sample evidence is strong enough for us to reject Ho. (c) Exponential smoothing method: method: Exponential smoothing is an averaging technique where the weightage given to the past data declines (an an exponential rate as the data recedes into the past. This all the values are taken into consideration, unlike in moving averages, where all data points prior to the period of the Moving average are ignored. If Ft is the one-period a head forecast made at time t and Dt is the demand for period t, then Ft = Ft-1 + α(Dt – Ft-1) = αDt + (1 – α) Ft-1 Where α is a soothing constant that lies between 0 1 but generally chosen values lie between 0.01 and 0.30
4
