Question

Data on pull-off force (pounds) for connectors used in an automobile engine application are as follows:

Answer 1

Answer:

a) 75.62

b) 75.2

Step-by-step explanation:

Data provided:

79.9,75.1,78.2,74.1,73.9,75.0,77.6,77.3,73.8,74.6,75.5,74.0,74.7,75.9,72.6,73.8,74.2,78.1,75.4,76.3,75.3,76.2,74.9,78.0,75.1,76.8

Sum = 1966.3

Total number of observations, n = 26

a) Mean is given as:

Mean = $\frac{\textup{Sum of all observations}}{\textup{Total number of observations}}$

or

Mean = $\frac{\textup{1966.3}}{\textup{26}}$

or

Mean = 75.62

b) For value that separates the weakest 50% of the connectors i.e median or the 50th percentile

Arranging the data in ascending order:

72.6, 73.8, 73.8, 73.9, 74, 74.1, 74.2, 74.6, 74.7, 74.9, 75, 75.1, 75.1, 75.3, 75.4, 75.5, 75.9, 76.2, 76.3, 76.8, 77.3, 77.6, 78, 78.1, 78.2, 79.9

i = $\frac{\textup{n}}{\textup{2}}$

or

i = $\frac{\textup{26}}{\textup{2}}$  = 13

When the number of observations is even the formula for median is:

$Median = \frac{\frac{n}{2}+\frac{n+2}{2}}{2}$

or

$Median = \frac{\frac{26}{2}+\frac{26+2}{2}}{2}$

or

$Median = \frac{13th+14th}{2}$

or

$Median = \frac{75.1+75.3}{2}$

or

Median = 75.2