A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16. a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams. b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams. c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.

a)0.099834

b) 0

Step-by-step explanation:

To solve for this question we would be using , z.score formula.

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

A candy maker produces mints that have a label weight of 20.4 grams. Assume that the distribution of the weights of these mints is normal with mean 21.37 and variance 0.16.

a) Find the probability that the weight of a single mint selected at random from the production line is less than 20.857 grams.

Standard Deviation = √variance

= √0.16 = 0.4

Standard deviation = 0.4

Mean = 21.37

x = 20.857

z = (x-μ)/σ

z = 20.857 - 21.37/0.4

z = -1.2825

P-value from Z-Table:

P(x<20.857) = 0.099834

b) During a shift, a sample of 100 mints is selected at random and weighed. Approximate the probability that in the selected sample there are at most 5 mints that weigh less than 20.857 grams.

z score formula used = (x-μ)/σ/√n

x = 20.857

Standard deviation = 0.4

Mean = 21.37

n = 100

z = 20.857 - 21.37/0.4/√100

= 20.857 - 21.37/ 0.4/10

= 20.857 - 21.37/ 0.04

= -12.825

P-value from Z-Table:

P(x<20.857) = 0

c) Find the approximate probability that the sample mean of the 100 mints selected is greater than 21.31 and less than 21.39.

25/6 hope it : )

step-by-step explanation: