Essay about Statistic: Normal Distribution and Mark Ans

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Assessment item 2—Assignment 2

|Due date: |4:00pm, Friday, Week 10 |ASSESSMENT |
|Weighting: |20% |2 |
|Length: |One file (.doc, .docx, or .rtf) no more than 5 MB | |

Question 1 4 Marks

The histogram below shows a sample of annual percentage returns on investment portfolios chosen by 60 investment managers in Australia. Note that the class interval 0 to less than 2 is represented by the class mid-point 1, and so on.

[pic]

For this sample,

(a) What is the mean annual return? 1 mark

Ans: mean = ∑x.fx = (1×10 + 3×15 + 5×20 + 7×10 + 9×5)/60 = 4.5

(b) What is the median annual return? 1 mark

Ans: median = 5 (the 30th ranked value)

(more detailed calculation: -

[pic]

(c) What is the standard deviation? 2 marks

Ans: standard deviation = square root of ∑(x- mean)2.fx = √5.42 = 2.327

(more detailed calculations:-

[pic] [pic] = 2.3470

Question 2 4 Marks

A researcher conducts a random sample survey of 5,000 Australian television viewers to determine which channel they watch most during weekdays between the evening hours 6:00pm and 9:00pm. The results are provided in the accompanying table.

|Channel |Number of Viewers |
|ABC | 750 |
|Seven | 1,050 |
|Nine | 900 |
|Ten | 800 |
|SBS | 850 |
|Others | 650 |

A surveyed viewer is chosen at random. Find the probability that, during the 6-9 pm time slot, the viewer

(a) Watches SBS 1 mark

Ans: P(SBS) = 850/5000 = 0.17

(b) Watches ABC or Channel Seven 1 mark

Ans: P(ABC or Channel Seven) = (750 + 1050)/5000 = 0.36

(c) Does not watch Channel Nine 1 mark

Ans: P(not Nine) = 1 – 900/5000 = 1 – 0.18 = 0.82

(d) Does not watch Channel Seven, Channel Nine or Channel Ten 1 mark

Ans: Required probability = 1 – (1050+900+800)/5000 = 1 – 0.55 = 0.45

Question 3 4 Marks

a) Assume that John has contracted to use a computer dating service. He knows that his personality, background, interest, and so on should make him initially compatible with 25% of the general population. If the
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