Lecture 6
Farmingdale State College
68.3%
of the datapoints (the X’s) will fall within 1 SD
of the mean.
95.4%
of the datapoints (the X’s) will fall within 2 SD
of the mean.
99.7%
of the datapoints (the X’s) will fall within 3 SD
of the mean.
Z-scores re-express the original data points (the X’s) in a way that intuitively lets us know:
How close the X is to the mean
Where it falls in the dispersion of the distribution
How much this participant is like the other person in the sample
Where it falls in the dispersion of the distribution
How different this participant is from the majority of people in the sample
\(z = 1 = 25 + 2 = 27\)
\(z = -2 = 25 - 2 = 23\)
\(z = 0.5 = 25 - .05 = 24.4\)
Mean-center the X values:
\(x - \bar{x}\)
Divide by the standard deviation
\(\frac{}{s}\)
The average grade on the exam was an 86 with a standard deviation of 4.
The average grade on the exam was an 86 with a standard deviation of 4. What is the probability of scoring a
94
or higher on the exam?
The average grade on the exam was an
86
with a standard deviation of4
. What is the probability of scoring a94
or higher on the exam?
\(z = \frac{x-\bar{x}}{s} = \frac{94-86}{4} = \frac{8}{4}=2\)
Z = 2
corresponds with 2.1%
and 0.1%
of the curve (2.2%
) chance.
The mean exam score was
86
with a standard deviation of4
. What is the probability of scoring between an82
and a90
?
The mean exam score was
86
with a standard deviation of4
. What is the probability of scoring between an82
and a90
?
What is the probability of scoring between an 82
and a 90
?
\(z = \frac{82-86}{4} = \frac{-4}{4} = -1\)
\(z = \frac{90-86}{4} = \frac{4}{4} = 1\)
There is a 68.3% chance of scoring between 82
and 90
.
Esmeralda’s stats professor tells her class that the average score on the exam was a 72 with a standard deviation of 6, and the distribution of scores was normal. Esmeralda wants to calculate the probability that she scored below
60
Esmeralda’s stats professor tells her class that the average score on the exam was a 72 with a standard deviation of 6, and the distribution of scores was normal. Esmeralda wants to calculate the probability that she scored below
60
Esmeralda wants to calculate the probability that she scored below 60
\(\frac{60-72}{6} = \frac{-12}{6} = -2\)
2.2%
chance that she failed.
What’s the chance she passed?
\(100-2.2 = 97.8%\)
In real life, we are often working with numbers with long decimal points rather than nice whole numbers.
Locate the whole number and first decimal point along the left side of the table.
Locate the second decimal point along the top of the table.
The Z-score Probability Table gives you the probability of that z-score or less.
.022
% chance of failing and and a 1-.022%
chance of passing.In a recent study on world happiness, participants were asked to evaluate their current lives on a scale from
0
to10
, where0
represents the worst possible life and10
represents the best possible life. The responses were normally distributed, with a mean of5.4
and a standard deviation of2.2
. Find the probability that a randomly selected study participant’s response was:
Less than 4
More than 8
Find the probability that a randomly selected study participant’s response was:
Less than 4
More than 8
Less than 4 (-.63)
Find .6 on the left size of the table (z column)
Find .03 column
Note where they intersect
Since we want the value less than 4, we need to subtract 1
Report the value as a percentage (\(z\times100\))
More than 8 (1.18)
Find 1.1 on the left size of the table (z column)
Find .08 column
Note where they intersect
Report the value as a percentage (\(z\times100\))
The scale of scores for an IQ test are approximately normal with mean
100
and standard deviation15
. The organization MENSA, which calls itself the “high IQ society”, requires a score of130
or higher.
What percent of adults would qualify for membership?
Find the probability of scores more than 130
\(z = \frac{x-\bar{x}}{s} = z = \frac{130-100}{15}=\frac{30}{15} = 2\)
More than 130 (2)
Find 2.0 on the left size of the table (z column)
Find .00 column
Note where they intersect
Remember we are thinking about 130 or more so we have to subtract 1 from our proportion.
Report the value as a percentage (\(z\times100\))