Question

The manager of a warehouse would like to know how many errors are made when a product’s serial number is read by a bar-code reader. Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors, per 1,000 scans each.

What is the mean and standard deviation for these six samples?

Answer 1

Answer with Step-by-step explanation:

Six samples are collected of the number of scanning errors: 36, 14, 21, 39, 11, and 2 errors

Mean=(Sum of all observations)/(Total number of observations)

=(36+14+21+39+11+2)/6

= 123/6

= 20.5

Standard deviation is the square root of mean of squares of deviation around mean

Deviation around mean:

36-20.5, 14-20.5, 21-20.5, 39-20.5, 11-20.5, and 2-20.5

15.5,-6.5,0.5,18.5,-9.5 and -18.5

Square of deviations:

240.25,42.25,0.25,342.25,90.25 and 342.25

Mean of square of deviations

=(240.25+42.25+0.25+342.25+90.25+342.25)/6

=176.25

square root of mean of deviations=
`√(176.25)=13.28`

Hence, Standard deviation=13.28

and Mean=20.5

Answer 2

Mean: 20.5

Standard Deviation: 11.5

The mean is the total of the numbers divided by the amount of numbers. So, add 36 + 14 + 21 + 39 + 11 + 2 to get 123. Now, divide 123 by 6 to find that the mean is 20.5.

The standard deviation is the mean of the distances from the numbers to the mean. So, find the distance from the mean for each number. You get 15.5, 6.5, 0.5, 18.5, 9.5, and 18.5. Find the mean of these distances. Start by adding them together to get 69, then divide that by 6 to get a standard deviation of 11.5.