A statistical quality control process for cereal production measures the weight of a cereal box. The population standard deviation is known to be 0.09 ounces. If we wish to estimate the population average cereal box weight to within 0.011 ounces and be 94% confident, how large a sample should be used

the sample should be 237

Step-by-step explanation:

The computation of the large the sample be used is shown below:

Given that

Standard deviation = 0.09 ounces

margin of error = 0.011

For 94% confidence interval, the z value be 1.88

As we know that

Margin of error = ((z score × standard deviation) ÷ (Sample size))^2

0.011 = (1.88 × 0.09) ÷ (Sample size))^2

0.011 = 0.1692 ÷ (Sample size))^2

So, the sample size be

= 15.38^2

= 236.60

Hence, the sample should be 237

-232

step-by-step explanation:

square root of 36/6 = 2.449489743

square root of 16/6 = 1.632993162

square root of 6 = 2.449489743

the answer is square root of 16/6 because the square root of 36/6 is the same answer as square root of 6 = 2.449489743