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Statistics, EXT1 S1 2020 HSC 12b*

When a particular biased coin is tossed, the probability of obtaining a head is `3/5`.

This coin is tossed 100 times.

Let `X` be the random variable representing the number of heads obtained. This random variable will have a binomial distribution.

  1. Find the expected value, `E(X)`.   (1 mark)

  2. By finding the variance, `text(Var)(X)`, show that the standard deviation of `X` is approximately 5.   (1 mark)

Show Answers Only

i.    `60`

ii.   `text(See Worked Solutions)`

Show Worked Solution

i.    `X = text(number of heads)`

`X\ ~\ text(Bin) (n, p)\ ~\ text(Bin) (100, 3/5)`

`E(X)= np= 100 xx 3/5= 60`
 

ii.   `text(Var)(X)= np(1-p)= 60 xx 2/5= 24`

`sigma(x)= sqrt24~~ 5`

Statistics, EXT1 S1 2025 HSC 4 MC

A Bernoulli random variable \(X\) has probability distribution

\(P(x)=\dfrac{x+1}{3}\)  for  \(x=0,1\).

What are the mean and variance of \(X\) ?

  1. \(E(X)=\dfrac{1}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{9}\)
  2. \(E(X)=\dfrac{1}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{3}\)
  3. \(E(X)=\dfrac{2}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{9}\)
  4. \(E(X)=\dfrac{2}{3}, \quad \operatorname{Var}(X)=\dfrac{2}{3}\)
Show Answers Only

\(C\)

Show Worked Solution

\(P(0)=\dfrac{1}{3}, \ P(1)=\dfrac{2}{3} \)

\(E(X) = \dfrac{1}{3} \times 0 + \dfrac{2}{3} \times 1 = \dfrac{2}{3}\)

\(E(X^2) = \dfrac{1}{3} \times 0^2 + \dfrac{2}{3} \times 1^2 = \dfrac{2}{3} \)

\(\text{Var}(X) = E(X^2)-E(X)^2 = \dfrac{2}{3}-\dfrac{4}{9}=\dfrac{2}{9} \)

\(\Rightarrow C\)

Statistics, EXT1 S1 2024 MET1 4

Let \(X\) be a binomial random variable where  \(X \sim \operatorname{Bi}\left(4, \dfrac{9}{10}\right)\).

  1. Find the standard deviation of \(X\).   (1 mark)

  2. Find  \(\operatorname{Pr}(X<2)\).   (2 marks)

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a.    \(\operatorname{sd}(X)=\dfrac{3}{5}\)

b.    \(\dfrac{37}{10\,000}\)

Show Worked Solution

a.     \(\operatorname{sd}(X)\) \(=\sqrt{np(1-p)}\)
    \(=\sqrt{4\times\dfrac{9}{10}\times\dfrac{1}{10}}\)
    \(=\sqrt{\dfrac{36}{100}}\)
    \(=\dfrac{3}{5}\)

 

b.     \(\operatorname{Pr}(X<2)\) \(=\operatorname{Pr}(X=0)+\operatorname{Pr}(X=1)\)
    \(=\ ^4C _0\left(\dfrac{9}{10}\right)^0\left(\dfrac{1}{10}\right)^4+\ ^4C_1\left(\dfrac{9}{10}\right)^1\left(\dfrac{1}{10}\right)^3\)
    \(= 1 \times \dfrac {1}{10\,000} + 4 \times \dfrac{9}{10} \times \dfrac{1}{1000}\)
    \(=\dfrac{37}{10\,000}\)
Mean mark (b) 51%.

Statistics, EXT1 S1 EQ-Bank 12

Four cards are placed face down on a table. The cards are made up of a Jack, Queen, King and Ace.

A gambler bets that she will choose the Queen in a random pick of one of the cards.

If this process is repeated 7 times, express the gambler's success as a Bernoulli random variable and calculate

  1. the mean.   (1 mark)

  2. the variance.   (1 mark)

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a.    `7/4`

b.    `21/16`

Show Worked Solution

a.    `text(Let)\ \ X = text(number of Queens chosen)`

`X\ ~\ text(Bin) (7,1/4)`

`E(X)` `=np`
  `= 7 xx 1/4`
  `=7/4`

 

b.    `text(Var)(X)` `= np(1-p)`
    `=7/4(1-1/4)`
    `= 21/16`

Statistics, EXT1 S1 EQ-Bank 1 MC

If `X` equals the number of successes in `n` independent Bernoulli trials, how many distinct values can `X` take?

  1. `\ n-1`
  2. `\ n(n-1)`
  3. `\ n`
  4. `\ n+1`
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`D`

Show Worked Solution

`text(Distinct values of)\ X\ text(in:)`

`text{1 trial = 2}\ \ (X=0 or 1)`

`text{2 trials = 3}\ \ (X=0, 1 or 2)`

 `vdots`

`n\ text{trials =}\ n+1\ \ (X=0, 1, …, n)`

`=>  D`

Statistics, EXT1 S1 EQ-Bank 13

On average, batsmen playing cricket in a T20 competition play a scoring shot two out of every three balls.

In a regular season, a total of 1200 overs that each contain six balls, are bowled.

Estimate how many overs would have at least five scoring shots.  (3 marks)

Show Answers Only

`421`

Show Worked Solution

`text(Let)\ \ X = text(number of scoring shots in a six ball over)`

`X\ ~\ text(Bin) (6, 2/3)`

`P(X >= 5)` `= P(X = 5) + P(X = 6)`
  `=\ ^6 C_5 ⋅ (2/3)^5 (1/3) + \ ^6 C_6 ⋅ (2/3)^6`
  `= 64/243 + 64/729`
  `= 256/729`

 

`text(Let)\ \ Y=\ text(number of overs with at least 5 scoring shots)`

`Y\ ~\ text(Bin)(1200, 256/729)`

`E(Y)` `=np`  
  `=1200 xx 256/729`  
  `=421.39…`  
  `=421\ \ text{(nearest over)}`  

Statistics, EXT1 S1 EQ-Bank 15

In an experiment, a pair of dice are rolled 70 times.

A success is recorded if the sum of the dice roll is 5 or less.

  1. What is the mean of this binomial distribution?

     

    Give your answer to one decimal place.   (3 marks)

  2. What is the standard deviation?

     

    Give your answer to one decimal place.   (1 mark)

Show Answers Only

a.    `19.4`

b.    `3.7`

Show Worked Solution

a.    `text(Array of possible roll totals:)`

STRATEGY: A table (or array) can be a very efficient and error minimising strategy in questions like this.

`P(5\ text(or less)) = 10/36 = 5/18`

`text(Let)\ X =\ text(number of rolls) <= 5`

`X\ ~\ text(Bin)(70, 5/18)`

`E(X)` `= np`
  `= 70 xx 5/18`
  `= 19.4`

 

b.     `text(Var)(X)` `= np(1-p)`
  `sigma^2` `= 70 xx 5/18(1-5/18)= 14.043`
  `:. sigma` `= 3.7\ \ (text(to 1 d.p.))`

Statistics, EXT1 S1 EQ-Bank 5 MC

When a standard 6-sided die is thrown, the probability that it shows a prime number is  `2/3`.

If 10 standard dice are thrown, the number, `N`, of times a prime number is showing has a binomial distribution.

What is the standard deviation of  `N`, correct to 3 decimal places?

  1. 0.222
  2. 0.471
  3. 1.491
  4. 2.222
Show Answers Only

`C`

Show Worked Solution

`N\ ~\ text(Bin)(n, p)\ ~\ text(Bin)(10, 2/3)`

`text(Var)(N)` `= np(1 – p)`
  `= 10 · 2/3(1 – 2/3)`
  `= 20/9`

 

`:. σ_N` `= sqrt(20/9)`
  `= 1.4907…`

 
`=>\ C`

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)

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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)

Show Answers Only

`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)

Show Answers Only

`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`