Aproduction 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

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 =

here,

z = 2.58 for 99% confidence interval

n = sample size

thus,

1 =

or

n = (2.58 × 5.5)²

or

n = 201.3561 ≈ 201

Hence,

option (c) n = 201

i think is g(1)=-1.

hope that i can

4 candies

step-by-step explanation:

7 x 28 = 196

200 - 196 = 4 candies remaining.