Discrete Probability Distributions Lefties Suppose a friend of yours told you th
ID: 3297113 • Letter: D
Question
Discrete Probability Distributions
Lefties
Suppose a friend of yours told you that 10% of people in the population are left-handed. You randomly select 15 people from BSB123 and count the number of lefties in the sample, denoted X. Answer the following questions assuming people are either left or right handed but not both.
Find out the probability distribution of X and present the probability distribution in the form of a table using 5 decimal points in your answers. (You have to use Excel to calculate the probabilities, as the binomial table only gives the probabilities correct to 3 d.p.)
(2 marks)
How many lefties would you expect the sample of 15 contains using
the formula of expected value of discrete random variable
the binomial formula for expected value
(Show working.)
(2 marks)
Explain what expectation in statistics means in the context of the Lefties example.
(2 marks)
What is the probability of getting 5 or more lefties in the sample of 15?
(Show working.)
(2 marks)
If the sample contains 5 lefties, would you suspect your friend might be wrong? Discuss fully.
(2 marks)
Explanation / Answer
Question 1)
Excel Formula:
x
p (x)
0
=BINOMDIST(0,15,0.1,0)
1
=BINOMDIST(1,15,0.1,0)
2
=BINOMDIST(2,15,0.1,0)
3
=BINOMDIST(3,15,0.1,0)
4
=BINOMDIST(4,15,0.1,0)
5
=BINOMDIST(5,15,0.1,0)
6
=BINOMDIST(6,15,0.1,0)
7
=BINOMDIST(7,15,0.1,0)
8
=BINOMDIST(8,15,0.1,0)
9
=BINOMDIST(9,15,0.1,0)
10
=BINOMDIST(10,15,0.1,0)
11
=BINOMDIST(D11,15,0.1,0)
12
=BINOMDIST(D12,15,0.1,0)
13
=BINOMDIST(D13,15,0.1,0)
14
=BINOMDIST(D14,15,0.1,0)
15
=BINOMDIST(D15,15,0.1,0)
x
p (x)
0
0.20589
1
0.34315
2
0.26690
3
0.12851
4
0.04284
5
0.01047
6
0.00194
7
0.00028
8
0.00003
9
0.00000
10
0.00000
11
0.00000
12
0.00000
13
0.00000
14
0.00000
15
0.00000
Question 2)
np = 15 * 0.10 = 1.5 ~ 2
Approximately 2 lefties would you expect the sample of 15 contains.
Question 3)
Here if we are going to select the 15 people from BSB123 then the average number of people who would be lefties is approximately 2.
Question 4)
P ( x >= 5 ) = 1 – P ( x < 5 ) = 1 – ( P ( x = 0 ) + P ( x = 1 ) + P (x = 2) + P ( x = 3 ) + P (x = 4 ) )
= 1 – (0.20589+ 0.34315+ 0.26690+ 0.12851+ 0.04284)
= 0.01271
The probability of getting 5 or more is 0.01271
Question 5)
The probabilities of 5 lefties is 0.01047 which is less than .05, therefore we can say that my friend would be wrong here.
x
p (x)
0
=BINOMDIST(0,15,0.1,0)
1
=BINOMDIST(1,15,0.1,0)
2
=BINOMDIST(2,15,0.1,0)
3
=BINOMDIST(3,15,0.1,0)
4
=BINOMDIST(4,15,0.1,0)
5
=BINOMDIST(5,15,0.1,0)
6
=BINOMDIST(6,15,0.1,0)
7
=BINOMDIST(7,15,0.1,0)
8
=BINOMDIST(8,15,0.1,0)
9
=BINOMDIST(9,15,0.1,0)
10
=BINOMDIST(10,15,0.1,0)
11
=BINOMDIST(D11,15,0.1,0)
12
=BINOMDIST(D12,15,0.1,0)
13
=BINOMDIST(D13,15,0.1,0)
14
=BINOMDIST(D14,15,0.1,0)
15
=BINOMDIST(D15,15,0.1,0)
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