Download Normal_Distribution pwr pt

Many thanks to the authors of IB Mathematical Studies Standard Level: For the IB diploma (International Baccalaureate) by
Peter Blythe, Jim Fensom, Jane Forrest and Paula Waldman De Tokman (26 Jul 2012).
Information from this book has been used in this PowerPoint.
The properties of a normal distribution:
•
•
•
•
•
•
•
•
•
It is a bell-shaped curve.
It is symmetrical about the mean, μ. (The mean, the mode and the median all
have the same value).
The x-axis is an asymptote to the curve.
The total area under the curve is 1 (or 100%).
50% of the area is to the left of the mean, and 50% to the right.
Approximately 68% of the area is within 1 standard deviation, σ, of the mean.
Approximately 95% of the area is within 2 standard deviations of the mean.
Approximately 99% of the area is within 3 standard deviations of the mean
The expected value is found by multiplying the number in the sample by the
probability.
The properties of a normal distribution:
•
•
•
•
•
It is a bell-shaped curve.
It is symmetrical about the mean, μ. (The mean, the mode and the median all have the same value).
The x-axis is an asymptote to the curve.
The total area under the curve is 1 (or 100%).
50% of the area is to the left of the mean, and 50% to the right.
μ
The properties of a normal distribution:
•
•
•
•
•
It is a bell-shaped curve.
It is symmetrical about the mean, μ. (The mean, the mode and the median all have the same value).
The x-axis is an asymptote to the curve.
The total area under the curve is 1 (or 100%).
50% of the area is to the left of the mean, and 50% to the right.
50%
50%
μ
The properties of a normal distribution:
•
•
•
•
•
•
It is a bell-shaped curve.
It is symmetrical about the mean, μ. (The mean, the mode and the median all have the same value).
The x-axis is an asymptote to the curve.
The total area under the curve is 1 (or 100%).
50% of the area is to the left of the mean, and 50% to the right.
Approximately 68% of the area is within 1 standard deviation, σ, of the mean.
68%
μ-σ
σ
σ
μ μ+σ
The properties of a normal distribution:
•
•
•
•
•
•
•
It is a bell-shaped curve.
It is symmetrical about the mean, μ. (The mean, the mode and the median all have the same value).
The x-axis is an asymptote to the curve.
The total area under the curve is 1 (or 100%).
50% of the area is to the left of the mean, and 50% to the right.
Approximately 68% of the area is within 1 standard deviation, σ, of the mean.
Approximately 95% of the area is within 2 standard deviations of the mean.
95%
σ
σ
σ
σ
μ - 2σ μ - σ μ μ + σ μ + 2σ
The properties of a normal distribution:
•
•
•
•
•
•
•
•
It is a bell-shaped curve.
It is symmetrical about the mean, μ. (The mean, the mode and the median all have the same value).
The x-axis is an asymptote to the curve.
The total area under the curve is 1 (or 100%).
50% of the area is to the left of the mean, and 50% to the right.
Approximately 68% of the area is within 1 standard deviation, σ, of the mean.
Approximately 95% of the area is within 2 standard deviations of the mean.
Approximately 99% of the area is within 3 standard deviations of the mean.
99%
σ
σ
σ
σ
σ
σ
μ - 3σ μ - 2σ μ - σ μ μ + σ μ + 2σ μ + 3σ
Example 1
The waiting times for a elevator are normally distributed with a mean of 1.5 minutes and a
standard distribution of 20 seconds.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the times within one, two and three standard deviations of the mean.
b) Find the probability that a person waits longer than 2 minutes 10 seconds for the elevator.
c) Find the probability that a person waits less than 1 minute 10 seconds for the elevator.
200 people are observed and the length of time they wait for an elevator is noted.
d) Calculate the number of people expected to wait less than 50 seconds for the elevator.
Example 1
The waiting times for a elevator are normally distributed with a mean of 1.5 minutes and a
standard distribution of 20 seconds.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the times within one, two and three standard deviations of the mean.
1.5 minutes = 90 seconds μ = mean = 90 seconds σ = standard deviation = 20 seconds
μ - 3σ
0
10
20
30
μ - 2σ
40
50
60
μ
μ-σ
70
80
90
μ+σ
μ + 2σ
μ + 3σ
100 110 120 130 140 150 160 170 180
Time (Seconds)
Example 1
The waiting times for a elevator are normally distributed with a mean of 1.5 minutes and a
standard distribution of 20 seconds.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the times within one, two and three standard deviations of the mean.
b) Find the probability that a person waits longer than 2 minutes 10 seconds for the elevator.
2 minutes 10 seconds = 130 seconds
2.5% or 0.025
Using symmetry about μ, and the fact that approximately 95% of the area is within
2 standard deviations of the mean, the two shaded areas together = 5%.
So the probability of waiting longer than 2 mins 10 secs is 2.5%, or 0.025.
μ - 3σ
0
10
20
30
μ - 2σ
40
50
60
μ
μ-σ
70
80
90
μ+σ
μ + 2σ
μ + 3σ
100 110 120 130 140 150 160 170 180
Time (Seconds)
Example 1
The waiting times for a elevator are normally distributed with a mean of 1.5 minutes and a
standard distribution of 20 seconds.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the times within one, two and three standard deviations of the mean.
b) Find the probability that a person waits longer than 2 minutes 10 seconds for the elevator.
2.5% or 0.025
c) Find the probability that a person waits less than 1 minute 10 seconds for the elevator.
1 minutes 10 seconds = 70 seconds
16% or 0.16
Using symmetry about μ, and the fact that approximately 68% of the area is within
2 standard deviations of the mean, the two shaded areas together = 32%.
μ
μ - 3σ
μ - 2σ μ - σ
μ + σ μ + 2σ μ + 3σ
0
10
20
30
40
50
60
70
80
90
100 110 120 130 140 150 160 170 180
Time (Seconds)
Example 1
The waiting times for a elevator are normally distributed with a mean of 1.5 minutes and a
standard distribution of 20 seconds.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the times within one, two and three standard deviations of the mean.
b) Find the probability that a person waits longer than 2 minutes 10 seconds for the elevator.
2.5% or 0.025
c) Find the probability that a person waits less than 1 minute 10 seconds for the elevator.
16% or 0.16
μ - 3σ
0
10
20
30
μ - 2σ
40
50
60
μ
μ-σ
70
80
90
μ+σ
μ + 2σ
μ + 3σ
100 110 120 130 140 150 160 170 180
Time (Seconds)
Example 1
The waiting times for a elevator are normally distributed with a mean of 1.5 minutes and a
standard distribution of 20 seconds.
200 people are observed and the length of time they wait for an elevator is noted.
d) Calculate the number of people expected to wait less than 50 seconds for the elevator.
We already know from part b) that the shaded area represents a probability of 2.5% or 0.025,
because 95% of the area is within 2 standard deviations of the mean.
So, the expected number of people is 2.5% of 200. 200 x 0.025 = 5
μ - 3σ
0
10
20
30
μ - 2σ
40
50
60
μ
μ-σ
70
80
90
μ+σ
μ + 2σ
5 people
μ + 3σ
100 110 120 130 140 150 160 170 180
Time (Seconds)
Example 2
The heights of 250 twenty-year-old women are normally distributed with a mean of 1.68m
and standard deviation of 0.06m.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the heights within one, two and three standard deviations of the mean.
b) Find the probability that a woman has a height between 1.56m and 1.74m.
c) Find the expected number of woman with a height greater than 1.8m.
Example 2
The heights of 250 twenty-year-old women are normally distributed with a mean of 1.68m
and standard deviation of 0.06m.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the heights within one, two and three standard deviations of the mean.
μ = mean = 1.68m
1.4
σ = standard deviation = 0.06m
μ - 3σ
μ - 2σ
μ–σ
μ
μ+σ
1.50
1.56
1.62
1.68
1.74
1.5
1.6
1.7
Height (m)
μ + 2σ μ + 3σ
1.80
1.8
1.86
1.9
2
Example 2
The heights of 250 twenty-year-old women are normally distributed with a mean of 1.68m
and standard deviation of 0.06m.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the heights within one, two and three standard deviations of the mean.
b) Find the probability that a woman has a height between 1.56m and 1.74m. 81.5%
The area between μ – 2σ and μ = 47.5% because 95% of the whole area is within
2 standard deviations of the mean. The area between μ and μ + σ= 34% because 68%
In fact later we will see
of the whole area is within 1 standard deviations of the mean.
that if we calculated this
Therefore the area μ – 2σ and μ + σ = 47.5% + 34% = 81.5%.
more accurately using
μ
μ - 3σ
μ - 2σ μ – σ
μ + σ μ + 2σ μ + 3σ the GDC, the value
1.50
1.4
1.5
1.56
1.62
1.6
1.68
1.7
Height (m)
1.74
1.80
1.8
1.86
would be 81.9% to 3 s.f.
1.9
2
Example 2
The heights of 250 twenty-year-old women are normally distributed with a mean of 1.68m
and standard deviation of 0.06m.
a) Sketch a normal distribution diagram to illustrate this information, indicating clearly the
mean and the heights within one, two and three standard deviations of the mean.
b) Find the probability that a woman has a height between 1.56m and 1.74m. 81.5%
6 people
c) Find the expected number of woman with a height greater than 1.8m.
We can see that the shaded area represents a probability of 2.5%, because 95% of the area is
within 2 standard deviations of the mean. So, the expected number of people is 2.5% of 250.
250 x 0.025 = 6
μ
μ - 3σ
μ - 2σ μ – σ
μ + σ μ + 2σ μ + 3σ
1.50
1.4
1.5
1.56
1.62
1.6
1.68
1.7
Height (m)
1.74
1.80
1.8
1.86
1.9
2
You can use the GDC to calculate values that are not whole multiples of the standard deviation
(or of course even if they are!)
Packets of coconut milk are advertised to contain 250ml. Akshat tests 75 packets.
He finds that the contents are normally distributed with a mean volume of 255ml and a
standard deviation of 8 ml.
μ
Find the probability that a packet contains more than 250ml.
255
First sketch a normal distribution diagram:
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
In this case: 250, 1E99, 255, 8)
Or: 250, 500, 255,8)
μ–σ
μ+σ
247
263
μ - 2σ
μ + 2σ
271 μ + 3σ
μ - 3σ 239
231
220
The solution on the calculator is 0.7340145322
So the probability that a packet contains more
than 250ml is 0.734 or 73.4%
230
279
240
250
260
270
280
You could just type in a very large/small
number (larger/smaller than μ +/- 3σ )
290
Example 3
The lifetime of a light bulb is normally distributed with a mean of 2800 hours and a standard
deviation of 450 hours.
a) Find the percentage of light bulbs that have a lifetime of less than 1950 hours. 2.95%
b) Find the percentage of light bulbs that have a lifetime of between 2300 and 3500 hours.
c) Find the percentage of light bulbs that have a lifetime of more than 3800 hours.
μ
Sketching the normal distribution diagram
2800
gives a clear idea of what is happening.
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
0
a) In this case: -1E99, 1950, 2800, 450)
Or: 0, 1950, 2800, 450)
The solution on the calculator is 0.0294532933
So the percentage of light bulbs that have a
lifetime of less than 1950 hours is 2.95%
You can see from the sketch that indeed the answer
should be a little more than 2.5%, because there
would be 2.5% with a lifetime of less than 1900 hours.
μ–σ
μ+σ
2350
3250
μ - 2σ
μ + 2σ
1900
3700
μ - 3σ
μ + 3σ
1450
1000
4150
2000
3000
4000
5000
Example 3
The lifetime of a light bulb is normally distributed with a mean of 2800 hours and a standard
deviation of 450 hours.
a) Find the percentage of light bulbs that have a lifetime of less than 1950 hours. 2.95%
b) Find the percentage of light bulbs that have a lifetime of between 2300 and 3500 hours.
c) Find the percentage of light bulbs that have a lifetime of more than 3800 hours. 80.7%
μ
Sketching the normal distribution diagram
2800
gives a clear idea of what is happening.
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
b) In this case: 2300, 3500, 2800, 450)
μ–σ
μ+σ
2350
3250
μ - 2σ
μ + 2σ
1900
3700
μ - 3σ
μ + 3σ
1450
0
1000
The solution on the calculator is 0.8068327753
So the percentage of light bulbs that have a
lifetime of between 2300 and 3500 hours is 80.7%
4150
2000
3000
4000
5000
Example 3
The lifetime of a light bulb is normally distributed with a mean of 2800 hours and a standard
deviation of 450 hours.
a) Find the percentage of light bulbs that have a lifetime of less than 1950 hours. 2.95%
b) Find the percentage of light bulbs that have a lifetime of between 2300 and 3500 hours.
c) Find the percentage of light bulbs that have a lifetime of more than 3800 hours. 80.7%
μ
1.31%
Sketching the normal distribution diagram
2800
gives a clear idea of what is happening.
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
c) In this case: 3800, 1E99, 2800, 450)
Or: 3800, 0, 2800, 450)
μ–σ
μ+σ
2350
3250
μ - 2σ
μ + 2σ
1900
3700
μ - 3σ
μ + 3σ
1450
0
The solution on the calculator is 0.0131341011
So the percentage of light bulbs that have a
lifetime of of more than 3800 hours is 1.31%
1000
4150
2000
3000
4000
5000
Example 3
The lifetime of a light bulb is normally distributed with a mean of 2800 hours and a standard
deviation of 450 hours.
a) Find the percentage of light bulbs that have a lifetime of less than 1950 hours. 2.95%
b) Find the percentage of light bulbs that have a lifetime of between 2300 and 3500 hours.
c) Find the percentage of light bulbs that have a lifetime of more than 3800 hours. 80.7%
μ
1.31%
120 light bulbs are tested.
2800
d) Find the expected number of light bulbs
with a lifetime of less than 2000 hours.
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
μ+σ
2350
3250
μ - 2σ
μ + 2σ
1900
3700
μ - 3σ
μ + 3σ
1450
0
d) In this case: -1E99, 2000, 2800, 450)
The solution on the calculator is 0.0377201305
So the percentage of light bulbs that have a
lifetime of more than 3800 hours is 3.77%
120 x 0.0377 = 4.5264….
So you would expect 4 or 5 light bulbs
μ–σ
1000
4150
2000
3000
4000
5000
Inverse normal calculations
Sometimes you are given the percentage area under the curve, i.e. the probability or the
proportion, and you are asked to find the value corresponding to it. This is called an inverse
normal calculation.
Always make a sketch to illustrate the information given.
You must always remember to use the area to the left
when using your GDC. If you are given the area to the
right of the value, you must subtract this from 1 (or 100%)
before using your GDC.
For example, an area of 5% above a certain value means
there is an area of 95% below it.
95%
5%
Example 4
The volume of cartons of milk is normally distributed with a mean of 995ml and a standard
deviation of 5ml.
It is known that 10% of the cartons have a volume of less than x. Find the value of x. 989ml
Sketching the normal distribution diagram gives
a clear idea of what is happening.
The shaded area represents 10% of the cartons.
Enter the percentage given as a decimal, the
mean, and the standard deviation, in order,
with commas between.
In this case: 0.1, 995, 5)
The solution on the calculator is 988.5922422
So x = 989ml (to 3 sig. fig.)
970
980
990
1000
1010
1020
Example 5
The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g.
a) Find the percentage of pears that weigh between 100g and 130g.
b) It is known that 8% of the pears weigh more than m g.
Find the value of m.
c) 250 pears are weighed. Calculate the expected number of pears that weigh less than 105g.
Sketch a diagram. μ = 110g
μ
σ = 8g
110
μ–σ
μ+σ
102
118
μ - 2σ
μ + 2σ
94
126
μ - 3σ
μ + 3σ
86
70
80
134
90
100
110
Weight (g)
120
130
140
150
Example 5
The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g.
a) Find the percentage of pears that weigh between 100g and 130g. 88.8%
Sketch a diagram. μ = 110g
μ
σ = 8g
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
110
μ–σ
μ+σ
102
118
μ - 2σ
μ + 2σ
94
126
μ - 3σ
μ + 3σ
86
70
80
134
90
100
110
120
Weight (g)
a) In this case: 100, 130, 110, 8)
The solution on the calculator is 0.8881404812
So the percentage of pears that weigh between 100g and 130g is 88.8%
130
140
150
Example 5
The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g.
a) Find the percentage of pears that weigh between 100g and 130g. 88.8%
Find the value of m. 121g
b) It is known that 8% of the pears weigh more than m g.
Sketch a diagram. μ = 110g
μ
σ = 8g
110
Enter the percentage given as a decimal, the
mean, and the standard deviation, in order,
with commas between.
μ–σ
μ+σ
102
118
μ - 2σ
μ + 2σ
94
126
μ - 3σ
μ + 3σ
86
70
80
134
90
100
110
120
130
Weight (g)
In this case the 8% of the pears are to the right of m, so 92% are to the left of m.
In this case: 0.92, 110, 8)
The solution on the calculator is 121.2405725
So m = 121g (to 3 sig. fig.)
140
150
Example 5
The weights of pears are normally distributed with a mean of 110g and a standard deviation of 8g.
a) Find the percentage of pears that weigh between 100g and 130g. 88.8%
Find the value of m. 121g
b) It is known that 8% of the pears weigh more than m g.
c) 250 pears are weighed. Calculate the expected number of pears that weigh less than 105g.
Sketch a diagram. μ = 110g
μ
σ = 8g
Enter the values:
Lower Bound,
Upper Bound, μ, σ
in order, with
commas between.
110
μ–σ
μ+σ
102
118
μ - 2σ
μ + 2σ
94
126
μ - 3σ
μ + 3σ
86
70
80
134
90
100
110
120
130
140
150
Weight (g)
a) In this case: -1E99, 105, 110, 8)
The solution on the calculator is 0.2659854678
So the percentage of pears that weigh between 100g and 130g is 26.6%.c
250 x 0.266 = 66.5, so the expected number of pears that weigh less than 105g is 66 or 67.