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Statistics, EXT1 S1 2012 MET2 3

Steve and Jess are two students who have agreed to take part in a psychology experiment. Each has to answer several sets of multiple-choice questions. Each set has the same number of questions, `n`, where `n` is a number greater than 20. For each question there are four possible options A, B, C or D, of which only one is correct.

  1. Steve decides to guess the answer to every question, so that for each question he chooses A, B, C or D at random.

     

    Let the random variable `X` be the number of questions that Steve answers correctly in a particular set.

    1. What is the probability that Steve will answer the first three questions of this set correctly?  (1 mark)

    2. Use the fact that the variance of `X` is `75/16` to show that the value of `n` is 25.  (1 mark)

  1. The probability that Jess will answer any question correctly, independently of her answer to any other question, is  `p\ (p > 0)`. Let the random variable `Y` be the number of questions that Jess answers correctly in any set of 25.

    If   `P(Y > 23) = 6 xx P(Y = 25)`, show that the value of  `p=5/6`.  (2 marks)

Show Answers Only

a.i.  `1/64`

a.ii.  `text(See Worked Solutions)`

b.  `text(See Worked Solutions)`

Show Worked Solution
a.i.    `Ptext{(3 correct in a row)}` `= (1/4)^3`
    `= 1/64`

 

a.ii.    `text(Var)(X)` `= np(1 – p)`
  `75/16` `= n(1/4)(3/4)`
  `75` `= 3n`
  `:. n` `= 25`

 

b.   `Y ∼\ text(Bin)(25,p)`

♦♦♦ Mean mark part (c) 19%.
`P(Y > 23)` `= 6xx P(Y = 25)`
`P(Y = 24) + P(Y = 25)` `= 6xx P(Y = 25)`
`P(Y = 24)` `= 5xx P(Y = 25)`
`((25),(24))p^24(1 – p)^1` `= 5p^25`
`25p^24(1 – p)` `= 5p^25`
`25p^24-25p^25-5p^25` `=0`
`25p^24-30p^25` `=0`
`5p^24(5 – 6p)` `= 0`

 
`:. p = 5/6,\ \ (p>0)\ \ text(… as required)`

Statistics, EXT1 S1 2017 MET2 18

Let  `X`  be a discrete random variable with binomial distribution  `X ~\ text(Bin)(n, p)`. The mean and the standard deviation of this distribution are equal.

Given that  `0 < p < 1`, what is the smallest number of trials, `n`, such that  `p ≤ 0.01`.   (2 marks)

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`99`

Show Worked Solution

`mu = np,\ \ text(Var)(X) = np(1-p) = σ_x ^2`

`text(Given)\ \ mu = σ_x,`

`np` `= sqrt(np(1 – p))`
`n^2p^2` `= np(1 – p)`
`np(np-1+p)` `=0`
`np-1+p` `=0,\ \ \ (np!=0)`
`p(n+1)` `=1`
`p` `=1/(n+1)`

 

`1/(n + 1)` `<= 1/100`
`n+1` `>=100`
`n` `>= 99`

 
`:. n_text(min) = 99`

Statistics, EXT1 S1 2015 MET2 10

The binomial random variable,  `X`, has  `E(X) = 2`  and  `text(Var)( X ) = 4/3.`

Calculate  `P(X = 1)`.   (3 marks)

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`64/243`

Show Worked Solution

`np = 2\ …\ (1)`

`np(1 – p) = 4/3\ …\ (2)`

`text(Solve simultaneous equations:)`

`text(Substitute)\ \ np=2\ \ text(into)\ (2)`

`2(1-p)` `=4/3`  
`2-2p` `=4/3`  
`p` `=1/3`  

 
`n = 6,quadp = 1/3`

`:. X ∼\ text(Bin)(6, 1/3)`

 

`:. P(X=1)` `= ((6),(1)) xx (1/3)^1 xx (2/3)^5`
  `= 6 xx 1/3 xx (2/3)^5`
  `=64/243`

Statistics, EXT1 S1 2008 MET2 5 MC

Let  `X`  be a discrete random variable with a binomial distribution. The mean of  `X`  is 1.2 and the variance of  `X`  is 0.72

The values of `n` (the number of independent trials) and `p` (the probability of success in each trial) are

A.   `n = 3,\ \ \ p = 0.6`

B.   `n = 2,\ \ \ p = 0.6`

C.   `n = 2,\ \ \ p = 0.4`

D.   `n = 3,\ \ \ p = 0.4`

Show Answers Only

`D`

Show Worked Solution

`X∼\ text(Bin) (n, p)`

`mu` `= 1.2`
`np` `= 1.2\ …\ (1)`

 

`text(Var) (X)` `= 0.72`
`np (1 – p)` `= 0.72\ …\ (2)`

 
`text(Solve simultaneous equations:)`

`:. n = 3,\ \ p = 0.4`

`=>   D`