Question

The amount of water in a bottle is approximately normally distributed with a mean of 2.40 liters with a standard deviation of 0.045 liter. Complete parts​ (a) through​ (e) below. a).What is the probability that an individual bottle contains less than 2.36 ​liters? b). If a sample of 4 bottles is​ selected, what is the probability that the sample mean amount contained is less than 2.36 ​liters? c).If a sample of 25 bottles is​ selected, what is the probability that the sample mean amount contained is less than 2.36 ​liters? d.) Explain the difference in the results of​ (a) and​ (c). Part​ (a) refers to an individual​ bottle, which can be thought of as a sample with sample size .1875 nothing. ​Therefore, the standard error of the mean for an individual bottle is 01 nothing times the standard error of the sample in​ (c) with sample size 25. This leads to a probability in part​ (a) that is ▼ the probability in part​ (c).

Answer 1

Solution:

We are given

Mean = µ = 2.40

SD = σ = 0.045

a).What is the probability that an individual bottle contains less than 2.36 ​liters?

We have to find P(X<2.36)

Z = (X - µ)/σ

Z = (2.36 - 2.40)/0.045

Z = -0.8888889

P(Z<-0.8888889) = 0.1870314

(by using z-table or excel)

P(X<2.36) = 0.1870314

Required probability = 0.1870314

b). If a sample of 4 bottles is​ selected, what is the probability that the sample mean amount contained is less than 2.36 ​liters?

Here, we have to find P(Xbar<2.36)

Z = (Xbar - µ)/[σ/sqrt(n)]

We are given n = 4

Z = (2.36 - 2.40)/(0.045/sqrt(4))

Z = -1.7777778

P(Z<-1.7777778) = 0.03772018

(by using z-table or excel)

Required probability = 0.03772018

c).If a sample of 25 bottles is​ selected, what is the probability that the sample mean amount contained is less than 2.36 ​liters?

Here, we have to find P(Xbar<2.36)

Z = (Xbar - µ)/[σ/sqrt(n)]

We are given n = 25

Z = (2.36 - 2.40)/(0.045/sqrt(25))

Z = -4.4444444

P(Z< -4.4444444) = 0.00000441

(by using z-table or excel)

Required probability = 0.00000441

d.) Explain the difference in the results of​ (a) and​ (c).

In part (a), we calculate the z score and probability for the individual score. In part (b), we calculate the z-score and probability for the sample mean score of group of individuals.