The sugar content in a one-cup serving of a certain

Solution 11E

Step1 of 6:

Let us consider a random variable X it represents the sugar content contained in a certain breakfast. Here random variable X follows normal distribution with mean standard deviation and n = 140.

That is

The probability density function of normal distribution is given by

, .

Where,

x = random variable

= mean of X

= variance of X

= standard deviation os X

= mathematical constant and its value is 3.14

Here our goal is:

a. We need to find a 95% confidence interval for the mean sugar content.

b. We need to find a 99% confidence interval for the mean sugar content.

c. We need to find confidence level of the interval (11.81, 11.99).

d. We need to find a 95% confidence interval specifies the mean to within ±0.1.

e. We need to find a 99% confidence interval specifies the mean to within ±0.1.

Step2 of 6:

a).

Here we have to find 95%  CI, let us take .

Now,

   = 0.025

Z-scores(are to the right) is given by

  is obtained from standard normal table(area under normal curve). In standard normal table we have to see where 0.9750 value falls, it falls in row 1.9 under column 0.06.

Hence,

  95% confidence interval for the mean sugar content is given by

(11.7177, 12.0822)

Hence, 95% confidence interval for the mean sugar content is (11.7177, 12.0822).