smc-638-35-Find n/p given E(X) Var(X)

Probability, MET2 2020 VCAA 8 MC

Items are packed in boxes of 25 and the mean number of defective items per box is 1.4

Assuming that the probability of an item being defective is binomially distributed, the probability that a box contains more than three defective items, correct to three decimal places, is

`0.037`
`0.048`
`0.056`
`0.114`
`0.162`

Show Answers Only

`B`

Show Worked Solution

`X\ ~\ text(Bi)(25, p)`

♦ Mean mark 50%.

`E(X)`	`=np = 1.4`
`25p`	`=1.4`
`p`	`=1.4/25 = 0.056`

`text(Pr)(X>3)`	`=\ text(Pr) (X>=4)`
	`=0.048\ \ text{(to 3 d.p.)}`

`=>B`

Statistics, MET2 2021 VCAA 12 MC

For a certain species of bird, the proportion of birds with a crest is known to be `3/5`.

Let `overset^P` be the random variable representing the proportion of birds with a crest in samples of size `n` for this specific bird.

The smallest sample size for which the standard deviation of `overset^P` is less than 0.08 is

Show Answers Only

`D`

Show Worked Solution

`text{Solve by CAS:}`

`sqrt{{0.6 xx 0.4}/{n}} < 0.08`

`n > 37.5`

`=> D`

Probability, MET2 2017 VCAA 18 MC

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

Given that `0 < p < 1`, the smallest number of trials, `n`, such that `p ≤ 0.01` is

`37`
`49`
`98`
`99`
`101`

Show Answers Only

`D`

Show Worked Solution

`E(X)`	`= 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)`

`:.\ text(If)\ \ 1/(n + 1)`	`<= 0.01`
`n`	`>= 99`

`=> D`

Probability, MET2 2008 VCAA 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 = 4,\ \ \ \ \ p = 0.3`

B. `n = 3,\ \ \ \ \ p = 0.6`

C. `n = 2,\ \ \ \ \ p = 0.6`

D. `n = 2,\ \ \ \ \ p = 0.4`

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

Show Answers Only

`E`

Show Worked Solution

`X∼\ text(Bi) (n, p)`

`text(E) (X)`	`= 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`

`=> E`

Probability, MET2 2012 VCAA 3

Steve, Katerina and Jess are three students who have agreed to take part in a psychology experiment. Each student is 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.

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. Find, to four decimal places, the probability that Steve will answer at least 10 of the first 20 questions of this set correctly. (2 marks)
3. Use the fact that the variance of `X` is `75/16` to show that the value of `n` is 25. (1 mark)

Part (b) is no longer in the syllabus and is removed.

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 `text(Pr) (Y > 23) = 6 text(Pr) (Y = 25)`, show that the value of `p` is `5/6`. (2 marks)
From these sets of 25 questions being completed by many students, it has been found that the time, in minutes, that any student takes to answer each set of 25 questions is another random variable, `W`, which is normally distributed with mean `a` and standard deviation `b`.
719
It turns out that, for Jess, `text(Pr)(Y >= 18) = text(Pr) (W >= 20)` and also `text(Pr)(Y >= 22) = text(Pr)(W >= 25)`.

Calculate the values of `a` and `b`, correct to three decimal places. (4 marks)

Show Answers Only

1. `1/64`
2. `0.0139`
3. `text(See Worked Solutions)`
`text(No longer in syllabus.)`
`text(See Worked Solutions)`
`a = 24.246, b = 2.500`

Show Worked Solution

a.i.	`text{Pr(3 correct in a row)}`	`= (1/4)^3`
		`= 1/64`

a.ii. `X =\ text(Number Steve gets correct)`

`X ∼\ text(Bi)(20,1/4)`

`text(Pr)(X >= 10) = 0.0139\ \ text{(4 d.p.)}`

`[text(CAS: binom Cdf)\ (20,1/4,10,20)]`

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

b.i. & b.ii. `text(No longer in syllabus.)`

c. `Y ∼\ text(Bi)(25,p)`

♦♦♦ Mean mark part (c) 19%.

`text(Pr)(Y > 23)`	`= 6text(Pr)(Y = 25)`
`text(Pr)(Y = 24) + text(Pr)(Y = 25)`	`= 6text(Pr)(Y = 25)`
`text(Pr)(Y = 24)`	`= 5 text(Pr)(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)`

d. `W ∼\ text(N)(a,b^2)`

♦♦♦ Mean mark part (d) 19%.

`text(Pr)(Y >= 18)`	`= text(Pr)(W >= 20)`
`0.9552…`	`= text(Pr)(W >= 20)`
`:. text(Pr)(W<20)`	`=1-0.9552…`
`text(Pr)(Z<(20-a)/b)`	`=0.0447…`
`(20-a)/b`	`=-1.698…\ \ (1)`

MARKER’S COMMENT: Many students didn’t use binomial calculations for `Y` here.

`text(Pr)(Y >= 22)`	`= text(Pr)(W >= 25)`
`0.3815…`	`= text(Pr)(W >= 25)`
`text(Pr)(W < 25)`	`=1-0.3815…`
`text(Pr)(Z<(25-a)/b)`	`=0.6184…`
`(25-a)/b`	`=0.30136…\ \ (2)`

`text{Solve (1) and (2) simultaneously:}`

`a = 24.246\ \ text{(3 d.p.)}, \ b = 2.500\ \ text{(3 d.p.)}`

Probability, MET2 2015 VCAA 10 MC

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

`text(Pr)(X = 1)` is equal to

A. `(1/3)^6`

B. `(2/3)^6`

C. `1/3 xx (2/3)^2`

D. `6 xx 1/3 xx (2/3)^5`

E. `6 xx 2/3 xx (1/3)^5`

Show Answers Only

`D`

Show Worked Solution

`np = 2\ …\ (1)`

`np(1 – p) = 4/3\ …\ (2)`

`text(Solve simultaneous equations:)`

`n = 6,quadp = 1/3`

`:. X ∼\ text(Bi)(6, 1/3)`

`:. text(Pr)(X=1)`	`= ((6),(1)) xx (1/3)^1 xx (2/3)^5`
	`= 6 xx 1/3 xx (2/3)^5`

`=> D`