Assume that process that makes the parts can be altered to adjust the overall mean of the reflectiveness score. Also assume that if an individual part has a score less than 85, then the part is given a passing quality score. The adjusted value of the true mean in population S that would result in approximately 80% of parts from that population being given a passing score is: a. greater than 80
b. less than 80
c. equal to 80

c. equal to 80

Step-by-step explanation:

Given that:

The population mean = 80

The standard deviation is = 6

The sample mean = 85

The objective is to find the probability that is less than 85

From the standard normal table

P(X <85) = 0.7977

P(X< 85) = 79.77%

P(X< 85) = 80%

Therefore, The adjusted value of the true mean in population S that would result in approximately 80% of parts from that population being given a passing score is:  equal to 80    .

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;