The candy produced by a company has a sugar level that is normally distributed with a mean of 16.8 grams and a standard deviation of 0.7 grams. The company takes readings of every 10th bar off the production line. The reading points are 17.3, 14.9, 18.3, 16.5, 16.1, 17.4, 18.4. Is the process in control or out of control and why?

It is not in control  because at least two of the  of the data points are more than 2 standard deviations from the mean are far apart.

18.3,18.4 > 18.2 Upper Control limit

14.9, > 15.4 Lower Control Limit

Step-by-step explanation:

Find the limits using μ ± 3σ  and μ ± 2σ

Using μ ± 3σ

Upper Control Limit = μ +  3σ = 16.8 + 3(0.7)= 16.8 + 2.1= 18.9

Lower Control Limit = μ - 3σ =16.8 - 3(0.7)= 16.8 - 2.1= 14.7

Using μ ± 2σ

Upper Control Limit = μ +  2σ = 16.8 + 2(0.7)= 16.8 + 1.4= 18.2

Lower Control Limit = μ - 2σ =16.8 - 2(0.7)= 16.8 - 1.4= 15.4

The data points are 17.3, 14.9, 18.3, 16.5, 16.1, 17.4, 18.4.

Rearranging these data points   14.9, 16.1, 16.5, 17.3, 17.4, 18.3,18.4.

We see from the above that it is out of control because the values jump above and below the mean.

It is not in control as two of the  of the data points are more than 2 standard deviations from the mean are far apart.

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;