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?

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 :

1A. Mean of six sample =

⇒

⇒ Mean = 20.5

Standard deviation of six sample =

⇒ 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:

⇒

⇒ Mean = 23.5

Standard deviation of all 12 samples =

⇒ 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.

i suppose it should be (y=2x-5)

step-by-step explanation: