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Probability, MET1 2019 VCAA 6

Fred owns a company that produces thousands of pegs each day. He randomly selects 41 pegs that are produced on one day and finds eight faulty pegs.

  1. What is the proportion of faulty pegs in this sample?  (1 mark) 
  2. Pegs are packed each day in boxes. Each box holds 12 pegs. Let  `hat P`  be the random variable that represents the proportion of faulty pegs in a box.

     

    The actual proportion of faulty pegs produced by the company each day is `1/6`.

     

    Find  `text(Pr)(hat P < 1/6)`. Express your answer in the form  `a(b)^n`, where  `a`  and  `b`  are positive rational numbers and  `n`  is a positive integer.  (2 marks)

Show Answers Only
  1. `8/41`
  2. `(17/6) ⋅ (5/6)^11`
Show Worked Solution

a.    `text(Proportion of faulty pegs) = 8/41`
 

b.   `hat P = x/n = 1/6`

`text(Given)\ \ n = 12`

`1/6 = X/12 \ => \ X = 2`

`X\ ~\ text(Bi) (12, 1/6)`

`text(Pr)(hat P < 1/6)` `= text(Pr)(X < 2)`
  `= text(Pr)(X = 0) + text(Pr)(X = 1)`
  `= \ ^12 C_0 * (5/6)^12 + \ ^12 C_1 ⋅ (1/6)(5/6)^11`
  `= (5/6)^11 (5/6 + 12/6)`
  `= (17/6) ⋅ (5/6)^11`

Probability, MET2 2017 VCAA 16 MC

For random samples of five Australians, `hatP` is the random variable that represents the proportion who live in a capital city.

Given that  `text(Pr) (hatP = 0) = 1/243`, then  `text(Pr)(hatP > 0.6)`, correct to the four decimal places, is

  1. `0.0453`
  2. `0.3209`
  3. `0.4609`
  4. `0.5390`
  5. `0.7901`
Show Answers Only

`C`

Show Worked Solution

`text(Let)\ X =\ text(number who live in a capital)`

`=>X ~\ text(Bi) (5, p)`

`text(Pr) (X = 0)` `=\ ^5C_0 * p^0 (1-p)^5`
  `= 1/243`
`(1 – p)^5` `= 1/243`
`:.p` `= 2/3`

 

`:.\ text(Pr)(hatP > 0.6)`

`= text(Pr)(X > 3), \ \ (X ~\ text(Bi) (5, 2/3))`

`= text(Pr)(X >= 4)`

`= 0.4609`

`=> C`