Download Which is more likely?

Example: Phone bills for residents of Cincinnati have a mean
of $64 and a standard deviation of $9, as shown in the graph
below. Random samples of 36 phone bills are drawn from this
population and the mean of each sample is determined. Find
the mean and standard error of the mean of the sampling
distribution.
What is the probability that a
randomly selected sample of 36
phone bills will have a mean
that falls between $61 and $67?
(Use the sampling mean and standard error)
61 − 64
𝑧1 =
= −2 → 0.0228 𝑜𝑛 𝑡𝑎𝑏𝑙𝑒
1.50
67 − 64
𝑧2 =
= 2 → 0.9772 𝑜𝑛 𝑡𝑎𝑏𝑙𝑒
1.5
𝟎. 𝟗𝟕𝟕𝟐 − 𝟎. 𝟎𝟐𝟐𝟖 = 𝟎. 𝟗𝟓𝟒𝟒
Probability and the Central Limit Theorem
Distribution must be approximately normally distributed and this is to
find the probability for the mean of a sample.
z-score method:
Calculator Method:
-Convert desired mean
value(s) to a z-score:
-Find sampling distribution
mean, x and the standard
error,   
z
x  x
x

x
(

n
)
-Look up z-score(s) on table
for the probability
x
n
-Use normalcdf function with
sampling mean and standard
error
If you are asked to find the probability of a single value then you
use the population mean and standard deviation.
Credit card balances are normally distributed, with a mean of
$2870 and a standard deviation of $900.
a) What is the probability that a randomly selected credit card
holder has a balance less than $2500?
𝒛=
𝟐𝟓𝟎𝟎 − 𝟐𝟖𝟕𝟎
= −𝟎. 𝟒𝟏 → 𝟎. 𝟑𝟒𝟎𝟗 𝒐𝒏 𝒕𝒂𝒃𝒍𝒆 = 𝟑𝟒. 𝟎𝟗%
𝟗𝟎𝟎
b) If you randomly select 25 credit card holders, what is the
probability that their mean credit card balance is less than
$2500?
𝝈𝒙 =
𝟗𝟎𝟎
𝟐𝟓
= 𝟏𝟖𝟎
𝟐𝟓𝟎𝟎 − 𝟐𝟖𝟕𝟎
𝒛=
= −𝟐. 𝟎𝟔 → 𝟎. 𝟎𝟏𝟗𝟕 𝒐𝒏 𝒕𝒂𝒃𝒍𝒆 = 𝟏. 𝟗𝟕%
𝟏𝟖𝟎
Which is more likely?
The heights of American women aged 20 – 29 are normally
distributed with a mean of 64 inches and a standard deviation
of 2.4 inches. Are you more likely to select randomly one
woman with a height more than 62 inches or are you more
likely to select a sample of 20 women with an average height
more than 62 inches?
Single:
𝑧=
62 −64
2.4
Sample: 𝜎𝑥 =
2.4
20
= −0.83 → 0.2033 𝑜𝑛 𝑡𝑎𝑏𝑙𝑒 = 20.33%
= 0.54 so z =
62 −64
0.54
= −3.70 → 0 𝑜𝑛 𝑡𝑎𝑏𝑙𝑒 = 0%