7.3 – Using tables to find the value of z given a probability
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Test Your Understanding
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Test Your Understanding
Exercise 1
Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples).
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S1: Chapter 9�The Normal Distribution
Teacher Guidance
Go >
Go >
Go >
Go >
What does it look like?
The following shows what the probability distribution might look like for a random variable X, if X is the height of a randomly chosen person.
Height in cm (x)
p(x)
180cm
We expect this ‘bell-curve’ shape, where we’re most likely to pick someone with a height around the mean of 180cm, with the probability diminishing symmetrically either side of the mean.
A variable with this kind of distribution is said to have a normal distribution.
What does it look like?
180cm
Normal Distribution Q & A
Q1
Q2
Q3
190cm
170cm
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Q4
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180cm
Normal Distribution Q & A
Q1
Q2
Q3
190cm
170cm
Q4
Notation
The random variable X...
...is distributed...
Example:
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Notation
The random variable X...
...is distributed...
Example:
Z value
🖉 The Z value is the number of standard deviations a value is above the mean.
100
Example
IQ | Z |
100 | 0 |
130 | 2 |
85 | -1 |
165 | 4.333 |
62.5 | -2.5 |
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p(x)
Z value
🖉 The Z value is the number of standard deviations a value is above the mean.
100
Example
IQ | Z |
100 | 0 |
130 | 2 |
85 | -1 |
165 | 4.333 |
62.5 | -2.5 |
p(x)
You may be wondering why we have to look up the values in a table, rather than being able to calculate it directly. The reason is that calculating the area under the graph involves integrating (see C2), but the probability function for the normal distribution (which you won’t see here) cannot be integrated!
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Z table
100
130
115
Expand
Minimise
Use of the z-table
Suppose we’ve already worked out the z value, i.e. the number of standard deviations above or below the mean.
Bro Tip: We can only use the z-table when:
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This is clear by symmetry.
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Bro Tip: Either changing the sign of changing the direction of the inequality does “1 –”. If we do both, they cancel out.
Use of the z-table
Suppose we’ve already worked out the z value, i.e. the number of standard deviations above or below the mean.
Bro Tip: We can only use the z-table when:
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This is clear by symmetry.
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Bro Tip: Either changing the sign of changing the direction of the inequality does “1 –”. If we do both, they cancel out.
Test Your Understanding
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Test Your Understanding
Exercise 1
Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples).
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Exercise 1
Determine the following probabilities, ensuring you show how you have manipulated your probabilities (as per the previous examples).
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‘Standardising’
Z world
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‘Standardising’
Z world
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Example
Here’s how they’d expect you to lay out your working in an exam:
The heights in a population are normally distributed with mean 160cm and standard deviation 10cm. Find the probability of a randomly chosen person having a height less than 180cm.
No marks attached with this, but good practice!
A statement of the problem.
M1 for “attempt to standardise”
Look up in z-table
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Example
Here’s how they’d expect you to lay out your working in an exam:
The heights in a population are normally distributed with mean 160cm and standard deviation 10cm. Find the probability of a randomly chosen person having a height less than 180cm.
No marks attached with this, but good practice!
A statement of the problem.
M1 for “attempt to standardise”
Look up in z-table
Test Your Understanding
Edexcel S1 May 2012
Edexcel S1 May 2013 (R)
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Test Your Understanding
Edexcel S1 May 2012
Edexcel S1 May 2013 (R)
Probabilities for Ranges
100
z=0
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112
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Probabilities for Ranges
100
z=0
96
112
Test Your Understanding
P(100 < X < 107.5) = P(Z < 0.5) – 0.5 = 0.1915
P(123 < X < 151) = P(1.53 < Z < 3.4) = P(Z < 3.4) – P(Z < 1.53) = 0.0627
P(70 < X < 90) = P(-2 < Z < -0.67) = (1 – 0.7486) – (1 – 0.9772) = 0.2286
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Test Your Understanding
P(100 < X < 107.5) = P(Z < 0.5) – 0.5 = 0.1915
P(123 < X < 151) = P(1.53 < Z < 3.4) = P(Z < 3.4) – P(Z < 1.53) = 0.0627
P(70 < X < 90) = P(-2 < Z < -0.67) = (1 – 0.7486) – (1 – 0.9772) = 0.2286
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Exercise 2
[Jan 2013 Q4a] The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours. Find P(L > 127). = 0.0359
[Jan 2012 Q7a] A manufacturer fills jars with coffee. The weight of coffee, W grams, in a jar can be modelled by a normal distribution with mean 232 grams and standard deviation 5 grams. Find P(W < 224). = 0.0548
[May 2011 Q4a] Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60.�A child from the school is selected at random. Find the probability that this child runs 100 m in less than 15 s.
= 0.2420
[Jan 2011 Q8a] The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random. Find the probability that this tin contains more than 168 g.
= 0.0548
[May 2010 Q7a] The distances travelled to work, D km, by the employees at a large company are normally distributed with D ~ N( 30, 82 ). Find the probability that a randomly selected employee has a journey to work of more than 20 km.
= 0.8944
[May 2009 Q8a,b] The lifetimes of bulbs used in a lamp are normally distributed. A company X sells bulbs with a mean lifetime of 850 hours and a standard deviation of 50 hours.�(a) Find the probability of a bulb, from company X, having a lifetime of less than 830 hours.
= 0.3446�(b) In a box of 500 bulbs, from company X, find the expected number having a lifetime of less than 830 hours. = 172.3
[Jan 2009 Q6a] The random variable X has a normal distribution with mean 30 and standard deviation 5.Find P(X < 39). = 0.9641
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(On provided sheet)
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Exercise 2
[Jan 2013 Q4a] The length of time, L hours, that a phone will work before it needs charging is normally distributed with a mean of 100 hours and a standard deviation of 15 hours. Find P(L > 127). = 0.0359
[Jan 2012 Q7a] A manufacturer fills jars with coffee. The weight of coffee, W grams, in a jar can be modelled by a normal distribution with mean 232 grams and standard deviation 5 grams. Find P(W < 224). = 0.0548
[May 2011 Q4a] Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60.�A child from the school is selected at random. Find the probability that this child runs 100 m in less than 15 s.
= 0.2420
[Jan 2011 Q8a] The weight, X grams, of soup put in a tin by machine A is normally distributed with a mean of 160 g and a standard deviation of 5 g. A tin is selected at random. Find the probability that this tin contains more than 168 g.
= 0.0548
[May 2010 Q7a] The distances travelled to work, D km, by the employees at a large company are normally distributed with D ~ N( 30, 82 ). Find the probability that a randomly selected employee has a journey to work of more than 20 km.
= 0.8944
[May 2009 Q8a,b] The lifetimes of bulbs used in a lamp are normally distributed. A company X sells bulbs with a mean lifetime of 850 hours and a standard deviation of 50 hours.�(a) Find the probability of a bulb, from company X, having a lifetime of less than 830 hours.
= 0.3446�(b) In a box of 500 bulbs, from company X, find the expected number having a lifetime of less than 830 hours. = 172.3
[Jan 2009 Q6a] The random variable X has a normal distribution with mean 30 and standard deviation 5.Find P(X < 39). = 0.9641
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(On provided sheet)
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Exercise 2
(On provided sheet)
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Exercise 2
(On provided sheet)
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The reverse: Finding the z-value for a probability
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For nice ‘round’ probabilities, we have to look in the second z-table. You’ll lose a mark otherwise.
The reverse: Finding the z-value for a probability
For nice ‘round’ probabilities, we have to look in the second z-table. You’ll lose a mark otherwise.
Dealing with two-ended inequalities
Sometimes the range is two ended.
These can be one of two possible forms in an exam:
Bro Tip: The key is to use a diagram (or otherwise) to convert to a single-ended inequality.
0.3
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Suitable Diagram
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0.7
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0.15
0.15
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Suitable Diagram
Dealing with two-ended inequalities
Sometimes the range is two ended.
These can be one of two possible forms in an exam:
Bro Tip: The key is to use a diagram (or otherwise) to convert to a single-ended inequality.
0.3
0.7
0.15
0.15
Test Your Understanding
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We’ll come back to this later…
Test Your Understanding
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We’ll come back to this later…
Where we are so far
Understand what a Z-value means and the formula to calculate it.
Calculate a probability of being above or below a particular value.
Calculate a z-value corresponding to a probability.
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✔
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COMING SOON!
COMING NOW!
“Find the IQ corresponding with the bottom 30% of the population.”
Solve more complex problems (e.g. involving conditional probabilities)
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Where we are so far
Understand what a Z-value means and the formula to calculate it.
Calculate a probability of being above or below a particular value.
Calculate a z-value corresponding to a probability.
✔
✔
✔
COMING SOON!
COMING NOW!
“Find the IQ corresponding with the bottom 30% of the population.”
Solve more complex problems (e.g. involving conditional probabilities)
Retrieving the original value
What IQ corresponds to the bottom 78% of the population?
State the problem in probabilistic terms.
Identify z value
(as we did in the previous lesson).
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0.78
Bro Tip: Draw a diagram for these types of questions if it helps.
z = 0.77
‘Standardise’.
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Retrieving the original value
What IQ corresponds to the bottom 78% of the population?
State the problem in probabilistic terms.
Identify z value
(as we did in the previous lesson).
0.78
Bro Tip: Draw a diagram for these types of questions if it helps.
z = 0.77
‘Standardise’.
Retrieving the original value
What IQ corresponds to the bottom 90% of the population?
‘Standardise’
Identify z value
z = 1.2816
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0.9
z = 1.2816
State the problem in probabilistic terms.
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Retrieving the original value
What IQ corresponds to the bottom 90% of the population?
‘Standardise’
Identify z value
z = 1.2816
0.9
z = 1.2816
State the problem in probabilistic terms.
Retrieving the original value
What IQ corresponds to the bottom 30% of the population?
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‘Standardise’
State the problem in probabilistic terms.
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Identify z value
Retrieving the original value
What IQ corresponds to the bottom 30% of the population?
‘Standardise’
State the problem in probabilistic terms.
Identify z value
Retrieving the original value
What IQ does 80% of the population have a value more than?
Convert back into an IQ.
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Identify z value
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‘Standardise’
State the problem in probabilistic terms.
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Retrieving the original value
What IQ does 80% of the population have a value more than?
Convert back into an IQ.
Identify z value
‘Standardise’
State the problem in probabilistic terms.
Test Your Understanding
[May 2008 Q7b] A packing plant fills bags with cement. The weight X kg of a bag of cement can be modelled by a normal distribution with mean 50 kg and standard deviation 2 kg.
Find the weight that is exceeded by 99% of the bags. (5)
[May 2011 Q4b] Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60.
On sports day the school awards certificates to the fastest 30% of the children in the 100 m race. Estimate, to 2 decimal places, the slowest time taken to run 100 m for which a child will be awarded a certificate. (4)
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Test Your Understanding
[May 2008 Q7b] A packing plant fills bags with cement. The weight X kg of a bag of cement can be modelled by a normal distribution with mean 50 kg and standard deviation 2 kg.
Find the weight that is exceeded by 99% of the bags. (5)
[May 2011 Q4b] Past records show that the times, in seconds, taken to run 100 m by children at a school can be modelled by a normal distribution with a mean of 16.12 and a standard deviation of 1.60.
On sports day the school awards certificates to the fastest 30% of the children in the 100 m race. Estimate, to 2 decimal places, the slowest time taken to run 100 m for which a child will be awarded a certificate. (4)
Additional Practice (Outside of Class)
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Additional Practice (Outside of Class)
Exercise 3a
(On provided sheet)
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Exercise 3a
(On provided sheet)
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Exercise 3a
(On provided sheet)
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Exercise 3a
(On provided sheet)
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Exercise 3a
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Exercise 3a
(On provided sheet)
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Harder reverse probability questions
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Harder reverse probability questions
Quartiles and Percentiles
i.e. 25% of people have an IQ less than the lower quartile.
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Quartiles and Percentiles
i.e. 25% of people have an IQ less than the lower quartile.
Test Your Understanding
Edexcel S1 May 2012
Edexcel S1 May 2010
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Test Your Understanding
Edexcel S1 May 2012
Edexcel S1 May 2010
Exercise 3b
(On provided sheet)
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First deal with P(X > 35) = 0.025
Next deal with P(X < 15) = 0.1469
We now have two simultaneous equations. Solving gives:
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First deal with P(X > 35) = 0.025
Next deal with P(X < 15) = 0.1469
We now have two simultaneous equations. Solving gives:
✔
🗶
🗶
Test your understanding
For the weights of a population of squirrels, which are normally distributed, Q1 = 0.55kg and Q3 = 0.73kg.
Find the standard deviation of their weights.
Due to symmetry, μ = (0.55 + 0.73)/2 = 0.64kg
If P(Z < z) = 0.75, then z = 0.67.
0.64 + 0.67σ = 0.73
σ = 0.134
🗶
σ = 0.114
σ = 0.124
σ = 0.134
σ = 0.144
Only 10% of maths teachers live more than 80 years. Triple that number live less than 75 years. Given that life expectancy of maths teachers is normally distributed, calculate the standard deviation and mean life expectancy.
✔
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σ = 2.77
σ = 2.78
σ = 79
σ = 2.80
✔
🗶
🗶
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μ = 76.15
μ = 76.25
μ = 76.35
μ = 76.45
RIP
A Ingall
He loved math
Exam Questions
Edexcel S1 May 2013 (R)
Edexcel S1 Jan 2011
Edexcel S1 Jan 2002
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Exam Questions
Edexcel S1 May 2013 (R)
Edexcel S1 Jan 2011
Edexcel S1 Jan 2002
Summary
A z-value is:
The number of standard deviations above the mean.
A z-table is:
A cumulative distribution function for a normal distribution with mean 0 and standard deviation 1.
P(IQ < 115) = 0.8413
We can treat quartiles and percentiles as probabilities.
For IQ, what is the 85th percentile?
100 + (1.04 x 15) = 115.6
We can form simultaneous equations to find the mean and standard deviation, given known values with their probabilities.
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P(a < X < b) = P(X < b) – P(X < a)
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A normal distribution is good for modelling data which:
tails off symmetrically about some mean/ has a bell-curve like distribution.
A
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We need to use the second z-table whenever:
we’re looking up the z value for certain ‘round’ probabilities.
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Summary
A z-value is:
The number of standard deviations above the mean.
A z-table is:
A cumulative distribution function for a normal distribution with mean 0 and standard deviation 1.
P(IQ < 115) = 0.8413
We can treat quartiles and percentiles as probabilities.
For IQ, what is the 85th percentile?
100 + (1.04 x 15) = 115.6
We can form simultaneous equations to find the mean and standard deviation, given known values with their probabilities.
B
C
D
G
H
F
P(a < X < b) = P(X < b) – P(X < a)
A normal distribution is good for modelling data which:
tails off symmetrically about some mean/ has a bell-curve like distribution.
A
E
We need to use the second z-table whenever:
we’re looking up the z value for certain ‘round’ probabilities.
Mixed Questions
Edexcel S1 June 2001
104