Question

A brand of contact lenses is advertised as lasting an average of 19 months with a standard deviation of 2 months. A consumer group doubts that the mean is that high and recruits 59 volunteers to test the contact lenses. Let X represent the mean time that the lenses last for these volunteers.

(a). Is X normally distributed? Explain your answer.



(b). Find the mean of the sampling distribution of X.



(c). Find the standard error of X.



(d). Write the R code to calculate the probability that X exceeds 21.



(e). Suppose R returns the value 0.027 for part (d). Interpret this value in the context of the problem

Answer 1

Solution:

Average (X) = 19 months

Sample standard deviation (s) = 2 months

Sample size (n) = 59

Since the sample size is greater than 30, we can use the central limit theorem to approximate the sampling distribution to normal distribution.

then sampling distribution of sample mean follow normal distribution with mean = 19 and variance = (2^2)/59 = 4/59

(a) -> Hence the X is normally distributed by using Central limit Theorem.

(b) -> Mean of the sampling distribution of X = 19 months

(c) -> The standard error of X = Standard devaiaton of sampling distribution

=

= 0.26038

(d) -> The R code to calculate the probability that X exceeds 21 is:

pnorm(21,mean = 19,sd = 0.26038,lower.tail = FALSE)

(e) ->

Interpretation: If the returns value is 0.027, then 2.7% of the sample have the mean lasting time of contact lense greater than 21 months.