Question 1.

A manufacturer produces shirts, finds that 0.1% of the shirts are defective. The shirts are packed in boxes contains 500 shirts. A shopkeeper buys 100 boxes from the producers. Using Poisson distribution, find how many boxes will contain:-

1. no defective, and 2. At least two defectives.

Solution:-

Given: - N = 100, n = 500, p probability of defective = 0.001, λ = np = 500 X 0.001 =0.5 and e= 2.7128let the random variable X denote the number of defective shirts in a box of 500.then by Poisson probability law, the probability of x defective shirts in a box is given by:

Hence in a consignment of 100 boxes, the frequency (number) of boxes containing x defective shirt is:

1. Number of boxes containing no defective shirts =

2. Number of boxes containing at least two defective shirts=100[P (X ≥2)] = 100{1-P(X=0) – P (X=1)}=100(1 – 0.6065 – 0.6065 X 0.5) = 100 X 0.9025 = 9

Question 2. Fit a least-squares line data by using

a) X as the independent variableb) X as dependent variable

Length X (in)

70 63 72 60 66 70 74 65 62 67 65 68

Weight Y (lb)

155 150 180 135 156 168 178 160 132 145 139 152

Solution:-a) Considering X as an independent variable, so that the least-square line can be written as

or

Where and .

Length X Weight Y70 155 3.2 0.8 2.56 10.24 0.6463 150 -3.8 -4.2 15.96 14.44 17.6472 180 5.2 25.8 134.16 27.04 656.6460 135 -6.8 -19.2 130.56 46.24 368.6466 156 -0.8 1.8 -1.44 0.64 3.2470 168 3.2 13.8 44.16 10.24 190.4474 178 7.2 23.8 171.36 51.84 566.4465 160 -1.8 5.8 -10.44 3.24 33.6462 132 -4.8 -22.2 106.56 23.04 492.8467 145 0.2 -9.2 -1.84 0.04 84.6465 139 -1.8 -15.2 27.36 3.24 168 152 1.2 -2.2 -2.64 1.44 4.84

and

The required least-square line is

or this equation is called the regression line of y on x and is used for estimating Y from given value of X.

b) if X is the dependent variable, the required line is

or this equation is called the regression line of X on Y and is used for estimating X from given value of Y.

SECOND METHOD

and

We may reduce suitable constants from X and Y. we choose to subtract 65 from X and 150 from Y. then the result can be as shown in table.

5 5 25 25 25-2 0 4 0 07 30 49 210 900-5 -15 25 75 2251 6 1 6 365 18 25 90 3249 28 81 252 7840 10 0 0 100-3 -18 9 54 3242 -5 4 -10 250 -11 0 0 1213 2 9 6 4

and

Since and , the regression equation are

and in agreement with first method.

Question 3:-

In a partially destroyed laboratory, record of an analysis of correlation data, the following results are legible:Variance , Regression equations : What are :

i) the mean value of X and Y,ii) the correlation coefficient between X and Y, andiii) the standard deviation of Y ?

Solution :-i) since both line of regression pass through the point we have:

Solving we get

ii) let be the line of regression of Y on X and X on Y respectively. these equations can be put in the form :

and

Regression coefficient of Y on X =

And Regression coefficient of X on Y =

Hence

iii) We have

hence

Question No: 4The factor that influence the breaking strength of a synthetic fibre are being studied. Four producyion machine and three operators are chosen and the experiment is run using a fibre from the same production batch. The results are as follows:

Operator Machine 1 Machine 2 Machine 3 Machine 41 109 110 108 1102 110 110 111 1143 116 112 114 120

Analyze the data and draw conclusions. Use

Solution:

Since there are two parameter so contingency method will be used.

Operator Machine 1 Machine 2 Machine 3 Machine 4 Total1 109 110 108 110 4372 110 110 111 114 4453 116 112 114 120 462Total 335 332 333 344 1344

Assuming the null hypothesis as H0 : the breaking strength is independent of machine and operator.

Machine/operator f 0 f e f 0- f e (f 0- f e)2/ f e

1,1 109 108.92 0.08 0.0000581,2 110 110.91 -0.91 0.00751,3 116 115.16 0.85 0.006272,1 110 107.94 2.06 0.03892,2 110 109.92 0.08 0.0000582,3 112 114.12 -2.12 0.03933,1 108 108.27 -0.27 0.000673,2 111 110.25 0.75 0.00513,3 114 114.46 -0.46 0.00184,1 110 111.85 -1.85 0.03054,2 114 113.89 0.15 0.000194.3 120 118.25 1.75 0.0286Totals 1344 1344.00 0.00 X 0 =0.1589

No. of degrees of freedom in contingency table = (no. of rows-1) X (no. of column - 1) = ( 3 – 1 ) X ( 4 – 1 ) = 2 X 3 = 6

Here

So, the f-table value for f 6, 0.05 = 12.69 ( X value)Since the calculated value is much lesser than table value. So, null hypothesis is true.

Question no. 5A process is controlled with a fraction of non-conforming control chart with three sigma limit, n = 400, UCL = 0.161, CL= 0.080 and LCL = 0.Find the equivalent control chart for the number of non conforming.

Solution:

For fraction chart is used.

So, and

Here, (given)The equivalent control chart for the number (proportion) of non-conforming np chart is used.So, And = = 7.36

Control limit for np chart; Therefore, UCL = 8 + 22.08 = 30.08LCL = 8 – 22.08 = - 14.08