Download Class notes - Nayland Maths

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Standard Normal Distribution
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What if we don’t have a ‘nice’ number of standard deviations from the mean?
Eg: A population of gummy bears has a mean of 3.4kg and standard deviation of
0.5kg
What % are less than 4.1kg?
How can we compare distributions with different means and standard deviations?
Use STANDARD normal distribution
Standard deviation = 1
Mean = 0
z
xx

34 kg
0
z
  Standard deviation
0
7 kg
x  mean
25 kg x
z = the number of standard deviations from the mean
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Z values
Theta Ex 24.04
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z = the number of standard deviations from the mean
For all z values (to 2dp) a table lists the areas under the standard normal curve
This gives the corresponding probability for each z value.
The probability is for the region from z back to the middle (0)
Look up the z value to get the probability
0
z
P(0 < z < 1.2) = 0.3849
P(0 < z < 1.5) =
P(0 < z < 0.8) =
P(0 < z < 2.1) =
P(z < 1.2) =
P( z < 1.8) =
P(z > 0.6) =
0
1.2
P(z < -1.2) =
-1.2 0
0 0.6
P( z < -2.3) =
P(-1.4 < z < 1.8) =
-1.4 0
1.8
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Converting To Standard Normal
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A population of snails have a mean weight of 54g (standard deviation of 8g)
(a) Find the probability a snail is over 60g
Steps:
1) Draw a distribution curve
54g 60g
2) Convert to a standard normal distribution
3) Look up ‘z’ to get probability
4) Add / subtract 0.5?
0
(b) Find the probability a snail is under 68g
WORKING
1) Question
2) ‘Z’ value
3) Probability
4) Answer
54g
z=
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Normal Distribution Example
. Theta Ex24.06
A fish farm has salmon with mean weight of 880g and standard deviation of 160g
a) What is the probability a fish is under 800g?
b) What % of fish are under 1kg?
c) From a population of 2600 fish, how many fish would be between 600g and
1200g?
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Normal Distribution: Practice
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Annual sunshine hours in Nelson are normally distributed with a mean of 2200hr
and a standard deviation of 35hr.
1) In a randomly chosen year:
a) What is the probability that the annual sunshine hours in Nelson are over 2150hr?
b) What is the chance that the annual sunshine hours in Nelson is less than 2100hr?
2) For how many of the past 100 years would we expect the number of sunshine
hours to have been between 2150 and 2250hr?
1a) The z-value for 2150 is given by
z
2150  2200
 1.429
35
Working
1)
2)
3)
4)
Question
‘Z’ value
Probability
Answer
P(x > 2150)
= P( z > - 1.429)
= 0.5 + 0.4235
= 0.9235
1b) The z-vale for 2100 is given by
z
2150
-1.429
2200
0
2235
The probability the annual sunshine
hours are over 2150 h is 0.9235.
2100  2200
 2.857
35
1)
2)
3)
4)
Question
‘Z’ value
Probability
Answer
P(x < 2100)
= P( z < - 2.857)
2100
2200
= 0.5 – 0.4984
-2.857
0
= 0.0016 The probability the annual sunshine hours
are less than 2100 h is 0.0016.
2.
1)
2)
3)
4)
The z-values for 2150 h and 2250 h are -1.429 and 1.429
Question
P(2150< x < 2250)
‘Z’ value
= P( -1.429 < x < 1.429)
Probability
= 0.4235 + 0.4235
Answer
= 0.8470
2150
-1.429
2200
0
2250
1.429
Therefore over the past 100 years we would expect there to have been 84.7 or 85
years with between 2150 and 2250 h of sunshine.