Aussie Maths & Science Teachers: Save your time with SmarterEd
A discrete random variable `X` has a binomial distribution with a probability of success of `p = 0.1` for `n` trials, where `n > 2`.
If the probability of obtaining at least two successes after `n` trials is at least 0.5, then the smallest possible value of `n` is
`C`
`text(Pr) (X ≥ 2) ≥ 0.5`
| `text(Pr) (X = 0)` | `=\ ^n C_0 xx 0.1^0 xx 0.9^n=0.9^n` |
| `text(Pr) (X = 1)` | `=\ ^n C_1 xx 0.1^1 xx 0.9^{n-1}=nxx0.1xx0.9^(n-1)` |
| `text(Pr) (X ≥ 2)` | `= 1 – (text(Pr) (X=0) + text(Pr) (X=1))` |
`text{Solve for} \ n \ text{(by CAS):}`
`1 – (0.9^n + n xx 0.1 xx 0.9^{n-1}) ≥ 0.5`
`n = 17`
`=> \ C`
In a large population of fish, the proportion of angel fish is `1/4`.
Let `hat P` be the random variable that represents the sample proportion of angel fish for samples of size `n` drawn from the population.
Find the smallest integer value of `n` such that the standard deviation of `hat P` is less than or equal to `1/100`. (2 marks)
`n_text(min) = 1875`
`text(Let)\ X =\ text(number of angel fish),\ X ~ text(Bi) (n, 1/4)`
| `text(sd)\ (hat P)` | `= sqrt((p(1 – p))/n)` |
| `= sqrt((1/4 xx 3/4)/n)` | |
| `= sqrt(3/(16n))` |
| `text(Solve:)\ \ \ sqrt(3/(16n))` | `<= 1/100` |
| `3/(16 n)` | `<= 1/(10\ 000)` |
| `(30\ 000)/16` | `<= n` |
| `:. n` | `>= 1875` |
`:. n_text(min) = 1875`