Weights of the vegetables in a field are normally distributed. from a sample carl cornfield determines the mean weight of a box of vegetables to be 180 oz. with a standard deviation of 8 oz. he wonders what percent of the vegetable boxes he has grouped for sale will have a weight between 169 oz. and 191 oz. carl decides to answer the following questions about the population of vegetables from these sample statistics. you will use some answers more than once. carl calculates the z-score corresponding to the weight 169 oz. (to the nearest tenth). using the table, carl sees the percentage associated with this z-score is

about 0.75 hours

step-by-step explanation:

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;