Aussie Maths & Science Teachers: Save your time with SmarterEd
A local council is proposing to ban dog-walking on the beach. It is known that the proportion of households that have a dog is \(\dfrac{7}{12}\).
The local council wishes to poll \(n\) households about this proposal.
Let \(\hat{p}\) be the random variable representing the proportion of households polled that have a dog.
What is the smallest sample size, \(n\), for which the standard deviation of \(\hat{p}\) is less than 0.06?
\(B\)
\(E(\hat{p}) = p = \dfrac{7}{12} \)
\(\sigma^{2} = \dfrac{p(1-p)}{n}\)
\(\text{Find}\ n\ \text{such that}\ \ \sigma \lt 0.06 :\)
| \(\Bigg( \dfrac{6}{100}\Bigg)^{2} \) | \( \gt \dfrac{ \frac{7}{12} \times \frac{5}{12}}{n} \) | |
| \(n\) | \( \gt \dfrac{35}{144} \times \dfrac{100^2}{6^2}\) | |
| \( \gt 67.5\) |
\(\Rightarrow B\)
An airline company that has empty seats on a flight is not maximising its profit.
An airline company has found that there is a probability of 5% that a passenger books a flight but misses it. The management of the airline company decides to allow for overbooking, which means selling more tickets than the number of seats available on each flight.
To protect their reputation, management makes the decision that no more than 1% of their flights should have more passengers showing up for the flight than available seats.
Given management's decision and using the attached normal distribution probability table to find a suitable approximation, find the maximum number of tickets that can be sold for a flight which has 350 seats. (4 marks)
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`358`
`text{Let}\ \ X=\ text{number of passengers taking a flight}`
`X ~ text{Bin}(n, 0.95)`
`E(X)=0.05,\ \ text{Var}(X)=n(0.95)(1-0.95)=0.0475n`
`X\ text{can be approximated by}\ \ Y ~ N(0.95n, 0.0475n)`
`text{Find}\ n\ text{such that}\ \ P(Y>350)=0.01:`
`text{Using the probability table}`
`=> ztext{-score of 2.33 corresponds to (closest) upper tail probability < 0.01}`
| `2.33` | `=(350-0.95n)/sqrt(0.0475n)` | |
| `2.33sqrt(0.0475)sqrtn` | `=350-0.95n` |
`0.95n+2.33sqrt(0.0475)sqrtn-350=0`
| `sqrtn` | `=(-2.33sqrt(0.0475)+-sqrt((2.33sqrt(0.0475))^2-4(0.95)(-350)))/(2(0.95))` | |
| `=18.9288…\ \ (n>0)` | ||
| `=(18.9288…)^2` | ||
| `~~358.30` |
`:.\ text{Maximum tickets that can be sold = 358}`
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.
Find the smallest sample size for which the standard deviation of `overset^p` is less than 0.08. (2 marks)
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`38`
`E(overset^p) = p = 0.6`
`text(Var)(overset^p) = (0.6(1 – 0.6))/n = 0.24/n`
`sigma(overset^p) = sqrt(0.24/n)`
`text(If)\ \ sigma(overset^p) < 0.08,`
| `sqrt(0.24/n)` | `< 0.08` |
| `0.0064n` | `> 0.024` |
| `n` | `> 0.24/0.0064` |
| `n` | `> 37.5` |
`:. text(Smallest)\ \ n = 38`
At a certain factory, the proportion of faulty items produced by a machine is `p = 3/500`, which is considered to be acceptable. To confirm that the machine is working to this standard, a sample of size `n` is taken and the sample proportion `overset^p` is calculated.
It is assumed that `overset^p` is approximately normally distributed with `mu = p` and `sigma^2 = (p(1 - p))/n`.
Production by this machine will be shut down if `overset^p >= 4/500`.
The sample size is to be chosen so that the chance of shutting down the machine unnecessarily is less than 2.5%.
Find the approximate sample size required, giving your answer to the nearest thousand. (3 marks)
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`6000`
`E(overset^p) = p = 3/500 = 0.006`
`P(overset^p >= 4/500) < 2.5text(%)`
`=> P(overset^p >= 4/500) = P(ztext(-score) <= 2)`
| `2 xx sigma` | `= 4/500 – E(overset^p)` |
| `= 4/500 – 3/500` | |
| `= 1/500` | |
| `sigma` | `= 1/1000` |
| `text(Using)\ \ sigma^2` | `= (p(1 – p))/n` |
| `0.001^2` | `= (0.006(0.994))/n` |
| `n` | `= (0.006(0.994))/(0.001^2)` |
| `= 5964` | |
| `~~ 6000` |
The proportion of a population that have brown eyes is 0.35.
Using `overset^p` to represent the sample proportion, calculate the smallest sample size, `n`, such that the standard deviation of `overset^p` is below 0.02. (2 marks)
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`569`
`E(overset^p) = p = 0.35`
`text(Var)(overset^p) = (0.35(1 – 0.35))/n = 0.2275/n`
`sigma(overset^p) = sqrt(0.2275/n)`
`text(If)\ \ sigma(overset^p) < 0.02,`
| `sqrt(0.2275/n)` | `< 0.02` |
| `0.0004n` | `> 0.0275` |
| `n` | `> 0.2275/0.0004` |
| `n` | `> 568.75` |
`:. text(Smallest)\ \ n = 569`
A manufacturer makes torches that have a probability of 0.03 of being defective.
Let `overset^p` be the random variable that represents the sample proportion of torches for samples of size `n` drawn from production.
Find the smallest integer value of `n` such that the standard deviation of `overset^p` is less than `1/50`. (2 marks)
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`73`
`text(Let)\ \ X =\ text(number of defective torches)`
`X\ ~\ text(Bin)(n, p)\ ~\ text(Bin)(n, 0.03)`
| `text(Var)(overset^p)` | `= (p(1 – p))/n` |
| `σ(overset^p)` | `= sqrt((0.03(0.97))/n)` |
| `= sqrt(291/(10\ 000n))` |
| `sqrt(291/(10\ 000n))` | `< 1/50` |
| `291/(10\ 000n)` | `< 1/2500` |
| `10\ 000n` | `> 2500 xx 291` |
| `n` | `> (2500 xx 291)/(10\ 000)` |
| `> 72.75` |
`:.n_text(min) = 73`
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)
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`1875`
`text(Let)\ \ X =\ text(number of angel fish)`
`X\ ~\ text(Bin) (n, 1/4)`
| `σ\ (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`