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vlada-n [284]
3 years ago
9

A production process fills containers by weight. Weights of containers are approximately normally distributed. Historically, the

standard deviation of weights is 5.5 ounces. (This standard deviation is therefore known.) A quality control expert selects n containers at random. How large a sample would be required in order for the 99% confidence interval for to have a length of 2 ounces?
a. n  15
b. n  16
c. n  201
d. n  226
Mathematics
1 answer:
forsale [732]3 years ago
5 0

Answer:

option (c) n = 201

Step-by-step explanation:

Data provided in the question:

Standard deviation, s = 5.5 ounce

Confidence level = 99%

Length of confidence interval = 2 ounces

Therefore,

margin of error, E = (Length of confidence interval ) ÷ 2

= 2 ÷ 2

= 1 ounce

Now,

E = \frac{zs}{\sqrt n}

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 = \frac{2.58\times5.5}{\sqrt n}

or

n = (2.58 × 5.5)²

or

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

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