#### Transcript “Random” is not a synonym for “haphazard” but a description of a

```Section 5.2
The Sampling Distribution of the Sample Mean
Let x be the mean of a SRS of size n from a
population with mean µ and standard deviation σ.
The mean and standard deviation of x are:
x  
x 

n
x  
x 

n
Notice that  x = µ, which is the mean of the
population.
This makes x an unbiased estimator
of the population mean.
x  
x 

n
The standard deviation depends on two things – the
standard deviation of the population, and also on
the sample size.
In order to make the variability smaller (and so
guarantee that our estimate is more likely to be close
to the truth) we should use a large sample.
Central Limit Theorem:
Draw a SRS of size n from any population with
mean µ and finite standard deviation σ.
If the population is exactly normal, then so is the
distribution of x .
If the population is not normal, then when n is
large, the sampling distribution of the sample mean
is approximately normal:
x is ≈ N (  ,  / n )
“Large” will depend on the shape of the
population, but we can use the rule of thumb that if
the population is not normal, then n should be at
least 30 in order to say the distribution of x is
approximately normal.
Suppose we have a normal population with mean
µ = 95 and σ = 27.
If we take a SRS of size n = 9, what, if anything,
can we say about the distribution of x ?
Because the pop’n is normal, the distribution of x
is also normal. Therefore, we can say that
x is N(µ, σ/√n) = N(95, 27/√9)
= N(95, 9).
x is N(µ, σ/√n) = N(95, 27/√9) = N(95, 9).
This allows us to find P(90 ≤ x ≤ 100), i.e., the probability
that the sample mean (when n is 9) will be within 5 units of
the population mean.
100  95
z
 0.56
27
9
P(90 ≤ x ≤ 100) = 0.4246 when n = 9.
z = – 0.56 and + 0.56
If we take a SRS of size n = 49, what,
if anything, can we say about the
distribution of x ?
x is N(µ, σ/√n) = N(95, 27/√49)
= N(95, 3.857).
Find P(90 ≤ x ≤ 100).
x is N(95, 3.857).
Find P(90 ≤ x ≤ 100).
100  95
z
 1.30
27
49
P(90 ≤ x ≤ 100) = 0.8064 when
n = 49.
z = – 1.30 and + 1.30
Why would you expect these results?
P(90 ≤ x ≤ 100) is about 42% when n = 9.
Yet P(90 ≤
x
≤ 100) is about 81% when n = 49.
n=9
n = 49
We increase our chances of getting a sample mean that is
“close” to the truth (in this case, within 5 units of the
population mean of 95) when we use a larger sample!
n=9
n = 49
According to the International Mass Retail
Association, girls aged 13 to 17 spend an average
of \$31.20 on shopping trips in a month, with a
standard deviation of \$8.27.
If 85 girls in that age category are
randomly selected, what is the probability that
their mean monthly shopping expense is
between \$30 and \$33?
µ = 31.20 and σ = 8.27
n = 85 We want P(30 < x < 33)
We do not know if the population is
normally distributed or not, but with a sample
size of 85, we know that x will
be approximately normal, with
 x = µ = 31.20  x = σ ÷√n = 8.27 /√85
zleft = (30 – 31.2) / (8.27 /√85) = – 1.34
zright = (33 – 31.2) / (8.27 /√85) = 2.01
Area
= 0.9778 – 0.0901
= 0.8877
```