Section 6.2: Sampling Distributions and The Central Limit Theorem
The mean of the sampling distribution is denoted by μx and equals the
mean of the population: μx=μ
The standard deviation of the sampling distribution, sometimes called the standard error, is denoted by σx and equals the
standard deviation of the population divided by the square root of the sample size: σx= σ/ sqr root(n)
Decide whether it is appropriate to use the normal distribution to find probabilities for X̄: 1. A simple random sample of size 45 will be drawn from a population with mean μ= 15 and standard deviation σ= 3.5. 2. A simple random sample of size 8 will be drawn from a normal population with mean μ= −60 and standard deviation σ= 5. 3. A simple random sample of size 42 will be drawn from a normal population with mean μ= −118 and standard deviation σ= 24.8. 4. A simple random sample of size 24 will be drawn from a population with mean μ= 35 and standard deviation σ= 1.2.
1. yes n>30 2. yes, ~normal 3. yes, ~normal 4. no, n<30 and don't know if ~normal
Consider the following probability histogram. Below are the probability histograms for the sampling distribution of x for samples of size 3, 10, and 30. Note that the shapes of the histograms begin to
approximate a normal curve as the sample size increases.
sampling distribution of X̄-
If several samples are drawn from a population, they are likely to have different values for .x Because the value of x varies each time a sample is drawn, x is a random variable. For each value of the random variable, x, we can compute a probability. The probability distribution of x
Central Limit Theorem
Let X̄ be the mean of a large (n> 30) simple random sample from a population with mean μ and standard deviation σ. Then X̄ has an approximately normal distribution, with mean μx=μ and standard deviation σx= σ/ sqr root(n).
The mean age of college students is μ= 25 years, with a standard deviation of σ = 9.5 years. A simple random sample of 125 students is drawn. a)What is the probability that the sample mean age is greater than 26 years? b)Find the 30th percentile of the sample mean X̄.
a) n>30, so ~normal, so use normalcdf -> P(X̄ >26)= normalcdf(26, inf, 25, 9.5/sqr root 125)= .1196 b) X̄= invNorm(.3, 25, 9.5/sqrr 125)= 24.5yrs
Hereford cattle are one of the most popular breeds of cattle. Based on data from the Hereford Cattle Society, the mean weight of a one-year-old Hereford bull is 1135 pounds, with a standard deviation of 97 pounds. Would it be unusual for the mean weight of 100 head of cattle to be less than 1100 pounds? Find the 35th percentile
a) P(X̄<1100)= normalcdf(-inf, 1100, 1135, 97/sqrr 100)= 1.542E-4= 0.002, yes its unusual P<0.05 b) X̄= invNorm(.35, 1135, 97/sqrr 100)= 1131lb
A population has mean μ= 10 and standard deviation σ= 8. A sample of size 50 is drawn. a. Find the probability that X̄ is greater than 11. b. Would it be unusual for X̄ to be less than 8? c. Find the 30th percentile of X̄
a) P(X̄>11)= normalcdf(11, inf, 10, 8/sqrr 50)= .1884 b) P(X̄<8)= normalcdf(-inf, 8,10,8/sqrr 50)= .03, yes, P<0.05 c) X̄= invNorm(.3, 10, 8/sqrr 50)= 9
Among students at a certain college, the mean number of hours of television watched per week is μ= 10.5, and the standard deviation is σ= 3.6. A simple random sample of 16 students is chosen for a study of viewing habits. Let x be the mean number of hours of TV watched by the sampled students. Find the mean μx and the standard deviation σx of X̄.
μ= 10.5, σ= 3.6, n=16 μx= 10.5hrs σx= 3.6/ sqr root(16)= .9hrs
