Ch 7.1 Central Limit Theorem for Sample Means

Ch 7.1 Central Limit Theorem for Sample Means

Sample distribution of sample mean:

When sample means \( \bar{x} \)  of same size n taken from the same population, the Sample means have the following behavior:

1)  If the population distribution of X is normal, the distribution of \( \bar{x} \) is always normal for all sample size n.

   

                                         Sampling distribution of x-bar
 

 2)  When population distribution of X is not normal, The sampling distribution x-bar tends to be a normal distribution. The distribution become closer to normal when sample size increase.

Activity to discover the Central Limit Theorem: 

https://stats.libretexts.org/Bookshelves/Ancillary_Materials/02%3A_Interactive_Statistics/15%3A_Discover_the_Central_Limit_Theorem_Activity

Central Limit Theorem for Sample Mean:

For all sample of the same size n with n > 30, the sampling distribution of \( \bar{x} \) can be approximated by a normal distribution with mean μ and standard deviation \( \sigma _{\bar{x}} = \frac{\sigma}{\sqrt{n}} \)

Note: -This applies to all distribution of x. If X is normally distributed, n > 30 is not needed. Any n will work.

      -The sample should be a Simple Random Sample.

 Central Limit Theorem:    \( \mu _{\bar{x}} = \mu  \),

                                       \( \sigma _{\bar{x}} = \frac{\sigma}{\sqrt{n} \)}    

Ex1 A standardized test with scores that are normally distributed with mean μ = 150 and standard deviation σ = 18. A class of 20 students take the test. The mean score \( \bar{x} \) of the 20 students are calculated.

a)  Is the distribution of mean score  \( \bar{x} \) of 20 students Normally distributed?

Ans: Yes because the original score is Normal.

a)  What is the mean and standard deviation of  ?

    Use Central Limit Theorem:  mean = 150,  SD = \( \frac{18}{\sqrt{20}}  \) ≈ 4.0249

b) Find the probability that a student’s score is greater than 160.

 Use Online Normal Calculator, Mean = 150, SD = 4.0249

c)  Find the probability that the mean score \( \bar{x} \) of 20 students is greater than 160.

 

    Click above, enter 160. Recalculate.  P( \( \bar{x} \) > 160) = 0.0065

 

Ex2: Coke cans are filled so that the actual amounts have a mean of 12 oz and a standard deviation of 0.11 oz. The distribution of amount of coke is unknown.

a)  Is the distribution of mean amount of coke in 36 cans normally distributed?

    Yes, because n > 30, according to CLT, \( \bar{x} \) will be normally distributed.

b) What is the mean and standard deviation of  \( \bar{x} \)?

    Ans: according to CLT:  \( \mu_ \bar{x} \) = 12, \(\sigma _{\bar{x}} \) = 0.11/√36 ≈ 0.01833

c)   Find the percent of individual coke with amount between 11.9 to 12.1 oz.

     Use online Normal Calculator:  Mean = 12, SD = 0.11

     Click between, enter 11.9 and 12.1, Recalculate.  P( 11.9 < x < 12.1 ) = 0.6367

      63.67% of coke have amount between 11.9 oz to 12.1 oz.

d)  Find the percent of mean amount of 36 coke with between 11.9 and 12.1 oz.

    Use online Normal Calculator:  Mean = 12, SD = 0.01833,

     Click between, enter 11.9 and 12.1, Recalculate.  P( 11.9 < \( \bar{x} \) < 12.1 ) = 1

     100% of mean amount of 36 coke is between 11.9 and 12.1 oz.

 

Ex3. Annual incomes are known to have a distribution that is skewed to the right. Assume that 20

workers’ mean incomes \( \bar{x} \)  are collected.

a)   Will the distribution of mean income \( \bar{x} \)  be normally distributed?

Ans: No, since X is not normal and n < 30, CLT does not apply,  \( \bar{x} \) may not be normally distributed.