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STAT175 ASSIGNMENT 1 Due Wednesday 5th September, 5pm This assignment is to be word processed and completed using this document. Handwritten responses will not be marked. Enter all answers in the red...

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STAT175 ASSIGNMENT 1
Due Wednesday 5th September, 5pm
This assignment is to be word processed and completed using this document. Handwritten responses will not be marked. Enter all answers in the red boxes and show any relevant working in the blue boxes marked ‘working’.
All probabilities should be rounded to four significant figures and all other values to two decimal places.
All work is to be a student’s own work. The assignment is to be submitted through Turnitin. The file upload link will be available in the Assessments folder in iLearn before the due date. Late penalties will apply as shown in the Unit Guide.
Question 1 (18 marks)
5Lotto is a weekly lotto draw which randomly selects 5 main winning numbers between 1 and 40 as well as a bonus number. For each ticket bought, players select 5 numbers between 1 and 40. Tickets cost $1 each and 80% of the money collected is allocated to prizes across five divisions as summarised in the following table.
Division    Requires    Share of Prize Pool
1    All 5 main winning numbers    40%
2    Any 4 main winning numbers and the bonus number    8%
3    Any 4 main winning numbers    10%
4    Any 3 main winning numbers and the bonus number    12%
5    Any 3 main winning numbers    30%
How many ways can you select 5 numbers out of 40 numbers?        
Complete the following table, showing the probability of winning a prize in each of the five divisions.Show all working in the space provided below.
Division    Probability of Winning
1     XXXXXXXXXX
2     XXXXXXXXXX
3     XXXXXXXXXX
4     XXXXXXXXXX
5     XXXXXXXXXX
Answered Same Day Sep 04, 2020 STAT175

Solution

Viswanathan answered on Sep 05 2020
148 Votes
Question 1
D
i) Jane bought 6 tickets
Explanation
6C5 = (6 * 5 * 4 * 3 * 2 * 1)/(5 * 4 * 3 * 2 * 1) = 6
Question 2
A)
Here, X represents the difference between the two numbers showing on the dice. Therefore, X takes the values 0, 1, 2, 3, 4 and 5
The probability mass function of X is
The probability function for the return of player after one roll is
B)
The probability distribution table is given below
    Difference, x
    Frequency
    Probability, P (x)
    0
    6
    0.167
    1
    10
    0.278
    2
    8
    0.222
    3
    6
    0.167
    4
    4
    0.111
    5
    2
    0.056
    Total
    36
    
Here, the expected return is
E (x) = x * P (x) = 0 * 0.167 + 1 * 0.278 + 2 * 0.222 + 3 * 0.167 + 4 * 0.111 + 5 * 0.056
= 1.94
E (x2) = x * P (x) = 02 * 0.167 + 12 * 0.278 + 22 * 0.222 + 32 * 0.167 + 42 * 0.111 + 52 * 0.056
= 5.833
Variance = E (x2) – [E (x)] 2 = 5.833 – 1.94 * 1.94 = 2.0525
Standard Deviation =...
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