A production process that fills 12-ounce cereal boxes is known to have a population standard deviation of 0.009 ounce. If a consumer protection agency would like to estimate the mean fill, in ounces, for 12-ounce cereal boxes with a confidence level of 92% and a margin of error of 0.001, what size sample must be used?

Given Information:  

Margin of error = 0.001

standard deviation = σ = 0.009

Confidence level = 92%

Required Information:

Population size = n = ?

Step-by-step explanation:

As we know

Margin of error = z*(σ/√n)

Re-arranging the equation to find n

√n = σz/margin of error

z-score corresponding to 92% confidence level = 1.72

√n = 0.009*1.72/0.001

√n = 15.48

square both sides

n = 15.48² = 239.63 = 240

n = 240 size sample should be used.

-362

step-by-step explanation:

(12)(4)+2(12)-34(12)+2(12)-3

  \/         \/         \/         \/

  48 +   24 -   408 + 24 - 3

      \   /               \   /

          72     -     432 - 3 = -363

dalila is incorrect. the whisker on the left is very long, going all the way out to 5. the whisker goes from 5 to 30, so it has a spread of 25. the rest of the box plot values are closer together, so 5 is an outlier