The weights of ice cream cartons are normally distributed with a mean weight of ounces and a standard deviation of ounce. (a) What is the probability that a randomly selected carton has a weight greater than ounces? (b) A sample of cartons is randomly selected. What is the probability that their mean weight is greater than ounces? (a) The probability is nothing. (Round to four decimal places as needed.) (b) The probability is nothing. (Round to four decimal places as needed.)

The answer is below

Step-by-step explanation:

The weights of ice cream cartons produced by a manufacturer are normally distributed with a mean weight of 10 ounces and a standard deviation of 0.5 ounce. (a) What is the probability that a randomly selected carton has a weight greater than 10.21 ounces? (b) You randomly select 25 cartons. What is the probability that their mean weight is greater than 10.21 ounces

Given that:

Mean (μ) = 10 ounces, standard deviation (σ) = 0.5 ounces.

The z score is used to determine by how many standard deviations the raw score is above or below the mean. The z score (z) is given by:

a) For x = 10.21:

From the normal distribution table, probability that a randomly selected carton has a weight greater than 10.21 ounces  = P(x > 10.21) = P(z > 0.42) = 1 - P(z < 0.42) = 1 - 0.6628 = 0.3372

b ) For x = 10.21 and n = 25

From the normal distribution table, probability that a randomly selected carton has a weight greater than 10.21 ounces  = P(x > 10.21) = P(z > 2.1) = 1 - P(z < 2.1) = 1 - 0.9826 = 0.0174

see attached

step-by-step explanation:

given the function as;

write the equation replacing f(x) with y

form   and fill a table containing values of y with corresponding values of x

you can take a range for   values of x as -3≤x≤3

solving for x=-3

for x=-2

repeat the same procedure to get the values of y for the remaining ,x=-1 to x=3

plot the values of y against values   of x and obtain the following graph