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Time between failures of wire-cut (in hours)

 Time between failures of wire-cut (in hours)

2 22 32 39 46 56 76 79 88 93

3 24 33 44 46 66 77 79 89 99

5 24 34 45 47 67 77 86 89 99

9 26 37 45 55 67 78 86 89 99

21 31 39 46 56 75 78 87 90 102

1. Calculate the mean, median, and mode of time between failures of wire-cuts .

2. The company would like to know by what time 10% (ten percentile or P10) and 90% (ninety percentile or P90) of the wire-cuts will fail.

3. Calculate the values of P25 and P75.

ANS:-

Mean :- The mean, often referred to as the average, is a statistical measure used to find the central or typical value of a set of numbers. To calculate the mean, you add up all the numbers in the set and then divide by the total number of values. The mean provides a single value that represents the “center” of the data set.

Median:- The median is another measure of central tendency used in statistics. It is the middle value in a data set when the values are arranged in ascending or descending order. If there is an even number of values, the median is the average of the two middle values. The median is less affected by extreme values (outliers) than the mean and can be a better representation of the “typical” value in skewed data.

Mode:-The mode is the value that appears most frequently in a data set. Unlike the mean and median, the mode is not concerned with the magnitude of values but rather with their frequency of occurrence. A data set can have one mode (unimodal), more than one mode (multimodal), or no mode if all values occur with the same frequency.

1.

Mean = 57.64, median = 56, and mode = 46, 89 and 99

Mean = (2 + 3 +5+ ….. + 99 + 102) / 50

median = 50/2=25

On the table 25th position is 56.

mode =46,89 and 99 because it repeat more than one time.

And it’s the multimode status.

2.

Note that the data in Table 2.4 is arranged in increasing order in columns. The position of P10 = 10 × (51)/100 = 5.1

We can round off 5.1 to its nearest integer which is 5.

The corresponding value from table is 21 (10 percentage of observations in Table 2.4 have a value of less than or equal to 21).

That is, by 21 hours, 10% of the wire-cuts will fail.

In asset management (and reliability theory), this value is called P10 life.

Instead of rounding the value obtained from Eq. (2.3), we can use the following approximation: Position corresponding to P10 = 10 × (51)/100 = 5.1 Value at 5th position is 21.

Value at position 5.1 is approximated as 21 + 0.1 × (value at 6th position — value at 5th position) = 21 + 0.1(1) = 21.1 Position corresponding to P90 = 90 × 51/100 = 45.9 The value at position 45 is 90 and the value at position 45.9 is 90 + 0.9 (value at 46th position — value at 45th position) = 90 + 0.9 × (3) = 92.7

That is, 90% of the wire-cuts will fail by 92.7 hours.

3.

Position corresponding to P25 (1st Quartile or Q1 ) = 25 × 51/100 = 12.75 Value at 12th position is 33, so P25 = 33 + 0.75 (value at 13th position — value at 12th position) = 33 + 0.75 (1) = 33.75 Position corresponding to P75 (3rd Quartile or Q3 ) = 75 × 51/100 = 38.25 Value at 38th position is 86, so P75 = 86 + 0.25 (value at 39th position — value at 38th position) = 86 + 0.25 (0) = 86

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