Statistics
Sampling Distributions
Chapter 6
Example Problems
Sample Proportions
Sample Proportions
• Answer: To find the sample proportions, look
at each pair of values. So we have
3,3
• The sample proportion would be the number
of odd values (since this is what the problem
asked) / total number of values = 2 odd values
/ 2 total values = 1
Sample Proportions
• We can continue this with
3, 4 would be 1 odd / 2 total = 1/2 or 0.5
3,11 would be 2 odd / 2 total = 1
• And so on….until we get
• 0 odds: 4,4
1 odd: 3,4; 4,3; 4,11; 11,4
2 odd: 3,3; 3,11; 11,3; 11,11
Sample Proportions
• Now to find the probability remember it is the
number of possible odd / total. So if you look at
the previous 0 there are 1 out of 9 total.
• Next, to find the probability that we would get
one odd and one even would be the second row
and there are 4 groups out of 9 total or 4/9
• And finally to find even,even we actually could
use the complement 1 – (odd,odd and odd,even)
= 1 – 1/9 – 4/9 = 9/9 – 1/9 – 4/9 = 4/9
Normal Distribution Example
• Question: Assume that women's heights are
normally distributed (this is the key to use the
normal z table) with a mean given by mean =
63.4 inches and a standard deviation = 1.8
inches.
a. If a woman is randomly selected, find
the probability that her height is between
62.9 inches and 63.9 inches.
Normal Distribution Example
• Answer: Find the z value with the formula z =
(x – mean) / standard deviation for the lower
value of x = 62.9 and upper value x = 63.9
– Low z = (62.9 – 63.4) / 1.8 = -0.28 (rounded to two
decimals)
– Upper z = (63.9 – 63.4) / 1.8 = 0.28 (rounded to
two decimal places)
Normal Distribution Example
• So “between” would be the area between z = 0.28 and z = 0.28
Normal Distribution Example
• The area from the bottom of the normal curve
up to z = -0.28 is found in the Negative z scores
table. Look up -0.2 in the row and .08 in the
column to get 0.3897.
• The area from the bottom of the normal curve
up to z = 0.28 is found in the Positive z scores
table. Look up 0.2 in the row and .08 in the
column to get 0.6103.
• Since this is the area between, if you subtract
these two values you get the area between
• 0.6103 - 0.3897 = .2206
Normal Distribution Example 2
• Question: Cans of a certain beverage are labeled
to indicate that they contain 20 oz. The amounts
in a sample of cans are measured and the sample
statistics are n = 42 and sample mean = 20.04 oz.
If the beverage cans are filled so that population
mean = 20.00 oz (as labeled) and the population
standard deviation is = 0.109 oz (based on the
sample results), find the probability that a sample
of 42 cans will have a mean of 20.04 oz or
greater. Do these results suggest that the
beverage cans are filled with an amount greater
than 20.00 oz?
Normal Distribution Example 2
• Answer: Since the sample size is greater than
20 we can approximate with a normal
distribution using the formula
z
x

n

20 . 04  20 . 00
0 . 109
42
 2 . 38
Normal Distribution Example 2
• We look up the z = 2.38 in the Normal Table to
get the area = 0.9913. However, we are looking
at the “amount greater than 20.00 oz” and thus
will be 1 – 0.9913 = 0.0087 which gives are
probability that the sample of 42 cans will have a
mean of 20.04 oz or greater.
• We look to compare with 0.05 (standard if not
given) and see that this probability 0.0087 being
way less certainly makes this particular event rate
if the true mean given is 20.00 oz.