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.
1A.] What is the mean and standard deviation for these six samples?
2A.] What number of errors is made by all scans, based on these six samples?

Just to be sure, the manager has six more samples taken:
33, 45, 34, 17, 1, and 29 errors, per 1,000 scans each
1B.] How do the mean and standard deviation change, based on all 12 samples?
2B.] How reasonable is it to expect that the small sample represents larger samples?

Answer 1

1A)  Mean=20.5    SD=176.25
2A) 17

Answer 2

Answer:

1A. 20.5

2A. 14.54

1B. 14.625

2B. quite good reasonable

Step-by-step explanation:

Mean is used to measure central tendency (i.e. representative of data) and standard deviation is use to measure dispersion of data. The formula use to calculate mean and variance is :

$Mean(bar{x}) = \dfrac{Sum of all the observations}{Total number of observation}$

$Standard deviation(\sigma) = \sqrt{\frac{1}{n} \sum_{i=1}^{n}(x_{i}-\bar{x})^2}$

1A. Mean of six sample =

$= \dfrac{Sum of all the observations}{Total number of observation}$

⇒ $\dfrac{36+14+21+39+11+2}{6}$

⇒ Mean = 20.5

Standard deviation of six sample =

$= \sqrt{\frac{1}{6}[ (36-20.5)^2+(14-20.5)^2+(21-20.5)^2+(39-20.5)^2+(11-20.5)^2+(2-20.5)^2}]$

⇒ σ = 14.54

2A. Total number of error = 36 + 14 + 21 + 39 + 11 + 2 = 123

Total number of error made by all scans is 123 error per 6000 scans.

1B. Mean of all 12 samples is:

$= \dfrac{Sum of all the observations}{Total number of observation}$

⇒ $\dfrac{36+14+21+39+11+2+33+45+34+17+1+29}{12}$

⇒ Mean = 23.5

Standard deviation of all 12 samples =

$= \sqrt{\frac{1}{12}[ (36-23.5)^2+(14-23.5)^2+(21-23.5)^2+(39-23.5)^2+(11-23.5)^2+(2-23.5)^2+(33-23.5)^2+(45-23.5)^2+(34-23.5)^2+(17-23.5)^2+(1-23.5)^2+(29-23.5)^2}]$

⇒ σ = 14.625

2B. Taking small sample instead of large sample can be quite risky sometimes as larger sample give us more accurate result than small sample.

But here we can take a small sample because the mean of both the size of the sample is near about.