Question
Times between successive crashes of a computer system were generated for a 6-month period and are given in increasing order as follows (time in hours):
1 10 20 30 40 52 63 70 80 90 100 102
130 140 190 210 266 310 530 590 640 1340
The parameter , mean
.
Use the Kolmogorov-Smirnov test to examine the goodness of fit of exponential distribution.
Answer :
Word Count : 878
_________ _________ __________ _________ _______ __________ _________ _______.
_____ ___ _________ _______ ____ _______ ________ _________ _________ _____.
_____ ___ _________ ________ _________.
_______ ________ ____ _______ ____ ______.
___ _______ _______ _______ _________ ________ ________.
____ _____ ___ _________ _____ _______ ____ ________ _______.
______ ___ ____ ________ __________.
_____ _____ ______ ______ _________ _______.
_____ ______ __________ ______ _________ ______ ______ _____ ________ _________ ______.
_________ ____ ______ _____ __________ _____ _____ ______ _________ _________ _______.
_________ _________ _____ _______ ____ ____.
_____ ________ _______ __________ ___ ____ _________ ___ _________ ______.
__________ _____ ____ _____ __________ ___ ____.
______ _________ ___ ____ ______ _________ ________ ________ _______ _____ ________ ___.
________ _________ ______ ____ __________ ____ _______ ____ ______ ______ ______.
_________ ___ _____ ___ __________ _______.
_________ ______ _____ ___ ______.
_________ ______ _________ ___ ______ ________.
___ __________ ____ ________ ________ _____ ____.
__________ _________ ________ _____ ________ ____ ________ _________ ___ ___.
_________ __________ _____ ___ _______ ________ ________ ____.
________ _________ _________ ______ _________ _________ _____ ________ ____.
_________ _________ ________ _____ _________ ______ ____ ___ ________.
_____ ____ _____ ________ ___ ________.
_______ __________ ________ _________ _____ _______.
_____ __________ ________ ___ _______ ________ __________ ____.
________ ______ _________ ____ _______ _____ ___ _______.
__________ __________ _________ __________ _____ ________ ____ ________ ______.
____ ______ _____ ____ ___ ______ __________ _______.
__________ _______ ________ _______ ______ __________.
_________ __________ _______ ____ _______ _____ __________ ______ ______ _____ ________ ______.
_____ _______ ______ ________ _________ ______.
_____ ___ _______ ______ ________ _______ __________ _______ ____ __________.
__________ __________ __________ ____ ____.
______ _____ ________ ______ _________ ________ _________ ___ _________ ______ _____ _____.
______ ______ ______ _______ __________ _________ ___ __________ _____.
_________ _______ _________ ________ _____ _____ ___ _________.
_____ ___ _____ __________ _______ __________.
______ _________ ___ _________ __________ ___ _________ __________.
_____ _______ _________ ____ ______ ___ ___.
___ ___ _______ ______ ________ _________ _____ _________ ___ __________ _____.
___ _______ _____ ____ ________ _____ ___ _______ _______ ____.
__________ __________ ________ __________ ___ _____ __________ ______.
______ ______ __________ __________ ________ __________.
______ __________ _______ ____ ___ ____ ___ _______ _____ _________.
__________ __________ _________ _________ ____ _______ ____ ________.
___ ________ __________ _______ _____ ______ _____ ______ ____.
________ ________ _________ _____ _________.
_______ _______ ____ __________ ___ __________.
___ _______ _________ ______ ____ __________ _____ _______ ______ __________.
________ _________ ___ _______ ____ __________ ____.
__________ ______ ___ ___ _______ __________ _________ ___.
________ ________ ___ _______ ________ _________ ____ _______ _________ ____.
______ __________ ______ _________ _________ ________.
________ _______ ______ __________ ______.
__________ ______ ____ ______ ___ _____ __________ __________ _______ ________ _________.
__________ _______ ____ ___ ____ ________.
____ _________ _____ __________ ________ ____ _____ ______ ____ __________ ____ ______.
________ __________ __________ _________ _________ _______ ______ ______.
_____ ________ ______ ______ __________ ______ ___ __________ ________ _________ ___.
__________ _____ ______ _______ ___.
____ ________ _________ ____ ______ ______ ________ ____ ____ ____.
___ ____ _________ _______ _______ _________ ____ _________.
______ ___ ___ _______ ___.
________ ___ _____ ________ _______ ____ ______ ____.
______ ________ _____ ________ ____ ______ __________ ___.
__________ ________ _____ ________ ________ ____ __________.
_____ _________ _____ ______ _______ ________ ______.
__________ ________ _________ __________ __________.
______ ___ ________ ____ ___ _______ ______ _____ __________ ___ ________ __________.
_______ __________ ______ _____ _____ _________ _________ _________.
____ ____ ___ _______ ____ _______ ______ ______ ________ _____ _______ __________.
________ ______ _________ ________ ______ _______.
_______ _______ ____ __________ ____ ________ ____ _________ ____.
________ _______ _______ ________ ________ ___.
_______ __________ ___ __________ ____ _______ _______.
____ _______ ___ ____ ________ _______ _____ ____ ______.
______ ____ ________ _______ _________ ______ ______ ______.
_______ __________ ___ ________ _____ _______ _______ ____ ___ ________ _________ _________.
_______ ________ ___ ______ ______ _____.
Skip in 10
We are asked to test the goodness of fit of an exponential distribution using the Kolmogorov-Smirnov (K–S) test. Let's solve this step by step. Given data (in hours): 1, 10, 20, 30, 40, 52, 63, 70, 80, 90, 100, 102, 130, 140, 190, 210, 266, 310, 530, 590, 640, 1340 Number of observations: ( n = 22 ) Parameter: ( a = 0.00435 ), mean ( 1/\alpha = 230 \text{ hrs} ) --- Step 1: Exponential CDF The exponential cumulative distribution function is: [ F(t) = 1 - e^{-\alpha t} ] Here, ( \alpha = 1/230 \approx 0.00435 ), which matches the given ( a ). --- Step 2: Compute empirical CDF values The empirical CDF at the ( i^\text{th} ) data point is: [ F_{\text{emp}}(t_i) = \frac{i}{n}, \quad i = 1, 2, \dots, 22 ] So: [ F_{\text{emp}} = \frac{1}{22}, \frac{2}{22}, \dots, \frac{22}{22} ] Numerical values: [ F_{\text{emp}} = 0.0455, 0.091, 0.136, 0.182, 0.227, 0.273, 0.318, 0.364, 0.409, 0.455, 0.500, 0.545, 0.591, 0.636, 0.682, 0.727, 0.773, 0.818, 0.864, 0.909, 0.955, 1.0 ] --- Step 3: Compute theoretical CDF values [ F(t_i) = 1 - e^{-0.00435 t_i} ] Calculating each (approximation to 4 decimals): 1. ( t = 1 ) → ( F(1) = 1 - e^{-0.00435} \approx 1 - 0.9957 = 0.0043 ) 2. ( t = 10 ) → ( F(10) = 1 - e^{-0.0435} \approx
__________ ___ ____ _____ _______ ___ ___._________ _________ __________ _________ _______ __________ _________ _______.
_____ ___ _________ _______ ____ _______ ________ _________ _________ _____.
_____ ___ _________ ________ _________.
_______ ________ ____ _______ ____ ______.
___ _______ _______ _______ _________ ________ ________.
____ _____ ___ _________ _____ _______ ____ ________ _______.
______ ___ ____ ________ __________.
_____ _____ ______ ______ _________ _______.
_____ ______ __________ ______ _________ ______ ______ _____ ________ _________ ______.
_________ ____ ______ _____ __________ _____ _____ ______ _________ _________ _______.
_________ _________ _____ _______ ____ ____.
_____ ________ _______ __________ ___ ____ _________ ___ _________ ______.
__________ _____ ____ _____ __________ ___ ____.
______ _________ ___ ____ ______ _________ ________ ________ _______ _____ ________ ___.
________ _________ ______ ____ __________ ____ _______ ____ ______ ______ ______.
_________ ___ _____ ___ __________ _______.
_________ ______ _____ ___ ______.
_________ ______ _________ ___ ______ ________.
___ __________ ____ ________ ________ _____ ____.
__________ _________ ________ _____ ________ ____ ________ _________ ___ ___.
_________ __________ _____ ___ _______ ________ ________ ____.
________ _________ _________ ______ _________ _________ _____ ________ ____.
_________ _________ ________ _____ _________ ______ ____ ___ ________.
_____ ____ _____ ________ ___ ________.
_______ __________ ________ _________ _____ _______.
_____ __________ ________ ___ _______ ________ __________ ____.
________ ______ _________ ____ _______ _____ ___ _______.
__________ __________ _________ __________ _____ ________ ____ ________ ______.
____ ______ _____ ____ ___ ______ __________ _______.
__________ _______ ________ _______ ______ __________.
_________ __________ _______ ____ _______ _____ __________ ______ ______ _____ ________ ______.
_____ _______ ______ ________ _________ ______.
_____ ___ _______ ______ ________ _______ __________ _______ ____ __________.
__________ __________ __________ ____ ____.
______ _____ ________ ______ _________ ________ _________ ___ _________ ______ _____ _____.
______ ______ ______ _______ __________ _________ ___ __________ _____.
_________ _______ _________ ________ _____ _____ ___ _________.
_____ ___ _____ __________ _______ __________.
______ _________ ___ _________ __________ ___ _________ __________.
_____ _______ _________ ____ ______ ___ ___.
___ ___ _______ ______ ________ _________ _____ _________ ___ __________ _____.
___ _______ _____ ____ ________ _____ ___ _______ _______ ____.
__________ __________ ________ __________ ___ _____ __________ ______.
______ ______ __________ __________ ________ __________.
______ __________ _______ ____ ___ ____ ___ _______ _____ _________.
__________ __________ _________ _________ ____ _______ ____ ________.
___ ________ __________ _______ _____ ______ _____ ______ ____.
________ ________ _________ _____ _________.
_______ _______ ____ __________ ___ __________.
___ _______ _________ ______ ____ __________ _____ _______ ______ __________.
________ _________ ___ _______ ____ __________ ____.
__________ ______ ___ ___ _______ __________ _________ ___.
________ ________ ___ _______ ________ _________ ____ _______ _________ ____.
______ __________ ______ _________ _________ ________.
________ _______ ______ __________ ______.
__________ ______ ____ ______ ___ _____ __________ __________ _______ ________ _________.
__________ _______ ____ ___ ____ ________.
____ _________ _____ __________ ________ ____ _____ ______ ____ __________ ____ ______.
________ __________ __________ _________ _________ _______ ______ ______.
_____ ________ ______ ______ __________ ______ ___ __________ ________ _________ ___.
__________ _____ ______ _______ ___.
____ ________ _________ ____ ______ ______ ________ ____ ____ ____.
___ ____ _________ _______ _______ _________ ____ _________.
______ ___ ___ _______ ___.
________ ___ _____ ________ _______ ____ ______ ____.
______ ________ _____ ________ ____ ______ __________ ___.
__________ ________ _____ ________ ________ ____ __________.
_____ _________ _____ ______ _______ ________ ______.
__________ ________ _________ __________ __________.
______ ___ ________ ____ ___ _______ ______ _____ __________ ___ ________ __________.
_______ __________ ______ _____ _____ _________ _________ _________.
____ ____ ___ _______ ____ _______ ______ ______ ________ _____ _______ __________.
________ ______ _________ ________ ______ _______.
_______ _______ ____ __________ ____ ________ ____ _________ ____.
________ _______ _______ ________ ________ ___.
_______ __________ ___ __________ ____ _______ _______.
____ _______ ___ ____ ________ _______ _____ ____ ______.
______ ____ ________ _______ _________ ______ ______ ______.
_______ __________ ___ ________ _____ _______ _______ ____ ___ ________ _________ _________.
_______ ________ ___ ______ ______ _____.
IGNOU ACADEMY
Your IGNOU journey, made easier.
- Solved Assignments
- Study Notes
- Guess Papers
- Question Papers
Continue Reading — It's Free
Sign in with Google to unlock more of this answer.
✓ Free access
✓ Unlock up to 60% of the answer
✓ No payment required
✓ Unlock up to 60% of the answer
✓ No payment required
Quick Google sign-in. No separate registration form.
Related Questions
To determine the yield rate of wheat in a district of Punjab, 6 groups of 6 plots each were constructed. The data are given in the following table:
| Plot No. | Group 1 | Group 2 | Group 3 | Group 4 | Group 5 | Group 6 |
|---|---|---|---|---|---|---|
| 1 | 8 | 6 | 18 | 13 | 17 | 12 |
| 2 | 13 | 5 | 8 | 7 | 15 | 15 |
| 3 | 11 | 16 | 6 | 13 | 10 | 11 |
| 4 | 26 | 5 | 10 | 6 | 21 | 17 |
| 5 | 13 | 16 | 16 | 7 | 20 | 8 |
| 6 | 31 | 5 | 20 | 2 | 25 | 10 |
Draw all possible samples of size 2 from the population [2, 3, 4] and verify that . Also find variance of
.
Times between successive crashes of a computer system were generated for a 6-month period and are given in increasing order as follows (time in hours):
1 10 20 30 40 52 63 70 80 90 100 102
130 140 190 210 266 310 530 590 640 1340
The parameter , mean
.
Use the Kolmogorov-Smirnov test to examine the goodness of fit of exponential distribution.
Generate a complete cycle for the LCG given below: , with
. A man tosses an unbiased coin ten times. Using the first ten random numbers generated above, obtain a sequence of heads and tails by taking Head (H) as
.
A sample of 60 students is to be drawn from a population consisting of 600 students belonging to two villages, A and B. The means and standard deviations of their marks are give below:
| Villages | Stratum sizes ($N_i$) | Means ($x_i$) | Standard deviations |
|---|---|---|---|
| Village A | 400 | 60 | 20 |
| Village B | 200 | 120 | 80 |
In the following data, two values are missing. Estimate these values by Yates method and analyse the data by suitable technique.
| Treatments | Blocks | ||
|---|---|---|---|
| I | II | III | |
| A | 12 | 14 | 12 |
| B | 10 | y | 8 |
| C | x | 15 | 10 |
The distribution function of Pareto distribution is given by ,
.
Given a , generate a random number from the above distribution, when
and
.
Suppose , then find x.