Answer:
a)0.099834
b) 0
Detailed explanation:
To find the solution to this problem, we will employ the z-score formula.
The z-score is calculated using the formula z = (x-μ)/σ, where x is the raw score, μ is the mean of the population, and σ is the standard deviation of the population.
A mint producer is making mints labeled to weigh 20.4 grams. Assume the weights of these mints are normally distributed with a mean of 21.37 and a variance of 0.16.
a) Determine the probability that a single mint randomly selected from the line weighs less than 20.857 grams.
Standard Deviation = √variance
= √0.16 = 0.4
So, standard deviation = 0.4
The mean = 21.37
x = 20.857
Therefore, z = (x-μ)/σ
z = (20.857 - 21.37)/0.4
z = -1.2825
The P-value from the Z-Table provides us with:
P(x<20.857) = 0.099834
b) If 100 mints are randomly selected and weighed during a shift, approximate the probability that at most 5 of them weigh less than 20.857 grams.
Utilizing the z-score formula as z = (x-μ)/σ/√n
Where x equals 20.857, the standard deviation equals 0.4, the mean equals 21.37, and n equals 100
z = (20.857 - 21.37)/(0.4/10)
= (20.857 - 21.37)/0.04
= -12.825
The P-value derived from the Z-Table yields:
P(x<20.857) = 0
c) Estimate the probability that the mean of the 100 mints selected is higher than 21.31 and lower than 21.39.
