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 13 20 23 31 35 40 43 48 49 58 62 66 67 69 71 73 77 78 79 82 85 86 89 91 93 97 99 101 105 106 106 112 117 124 135 139 141 147 159 161 168 183 207 249 262 289 323 388 513 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

Answer:

Sample mean =119.42

Median = 92

25% trimmed mean = 102.42

10% trimmed mean = 95.69

Step-by-step explanation:

Data in increasing order :

12 13 20 23 31 35 40 43 48 49 58 62 66 67 69 71 73 77 78 79 82 85 86 89 91 93 97 99 101 105 106 106 112 117 124 135 139 141 147 159 161 168 183 207 249 262 289 323 388 513

Total no. of observations = 50

Sample mean = $\frac{\text{Sum of all observations}}{\text{Total number of observations}}$

= $\frac{5971}{50}$ = 119.42

Median: Since we have even number of observation

Median = $\frac{1}{2}({25\text{th term} + 26\text{th term})$ = $\frac{91+93}{2}$ = 92

10% Trimmed Mean: We remove 5 values from each side

Trimmed set = 35 40 43 48 49 58 62 66 67 69 71 73 77 78 79 82 85 86 89 91 93 97 99 101 105 106 106 112 117 124 135 139 141 147 159 161 168 183 207 249

Trimmed mean =  $\frac{\text{Sum of Trimmed Set}}{\text{Total number of observations}}$ = $\frac{4097}{40}$ = 102.42

25% Trimmed Mean: We remove 12 values from each side.

Trimmed set =  66 67 69 71 73 77 78 79 82 85 86 89 91 93 97 99 101 105 106 106 112 117 124 135 139 141

Trimmed mean =  $\frac{\text{Sum of Trimmed Set}}{\text{Total number of observations}}$ = $\frac{2488}{26}$ = 95.69