Question

A certain article reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. 12 14 20 23 32 37 40 43 48 51 58 61 64 66 69 70 75 77 78 79 81 83 86 90 91 93 96 100 102 105 106 106 113 119 124 135 138 141 147 157 162 169 184 206 247 264 289 323 389 512 Compute the sample median, 25% trimmed mean, 10% trimmed mean, and sample mean for the lifetime data. (Round your answers to two decimal places.) sample median 25% trimmed mean 10% trimmed mean sample mean

Answer 1

sample size, N = 50

a) Median = mean( Nth and (N+1)th value )

Nth value = 91

(N+1)th value = 93

Median = ( 91+93 )/ 2 = 184/2 = 92

b) 25% trimmed mean

Trimmed mean percent = 25/100 = 0.25

sample size, N = 50

To obtain trimmed count

g = Floor( Trimmed mean percent * sample size )

g = Floor( 0.25*50 ) = Floor(12.5) = 12

Remove 12 number from begining and 12 from end

Numbers left : 64 66 69 70 75 77 78 79 81 83 86 90 91 93 96 100 102 105 106 106 113 119 124 135 138 141

calculate mean of these

25% trimmed mean = 2487/26 = 95.65

c) 10% trimmed mean

Trimmed mean percent = 10/100 = 0.1

sample size, N = 50

To obtain trimmed count

g = Floor( Trimmed mean percent * sample size )

g = Floor( 0.1*50 ) = Floor(5) = 5

Remove 5 number from begining and 5 from end

Numbers left :  37 40 43 48 51 58 61 64 66 69 70 75 77 78 79 81 83 86 90 91 93 96 100 102 105 106 106 113 119 124 135 138 141 147 157 162 169 184 206 247

calculate mean of these

10% trimmed mean = 4097/ 40 = 102.425

d) Sample mean = 5975/50 = 119.5