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Statistics, EXT1 S1 2024 HSC 12c

A charity employs a worker to collect donations. There is a 0.31 chance that when the charity worker talks to someone a donation is made to the charity.

Each day the charity worker must talk to exactly 100 people.

Use a standard normal distribution table to approximate the probability that, on a particular day, at least 35% of the people talked to made a donation.   (3 marks)

Show Answers Only

\(P(\hat{p} \geqslant 0.35) =0.1922\)

Show Worked Solution

\(E(\hat{p})=p=0.31, n=100\)

\(\sigma^2=\dfrac{p(1-p)}{n}=\dfrac{0.31 \times 0.69}{100}=0.002139\)

\(\hat{p} \sim N\left(\mu, \sigma^2\right) \sim N(0.31,0.002139)\)

  \(P(\hat{p} \geqslant 0.35)\) \(=P\left(Z \geqslant \dfrac{x-\mu}{\sigma}\right)\)
    \(=P\left(Z \geqslant \dfrac{0.35-0.31}{\sqrt{0.002139}}\right)\)
    \(=P(Z \geqslant 0.87)\)
    \(=1-P(Z<0.87)\)
    \(=1-0.8078 \text { (from table)}\) 
    \(=0.1922\)

Statistics, EXT1 S1 2023 HSC 11f

A recent census found that 30% of Australians were born overseas.

A sample of 900 randomly selected Australians was surveyed.

Let \(\hat{p}\) be the sample proportion of surveyed people who were born overseas.

A normal distribution is to be used to approximate \(P(\hat{p} \leq 0.31)\).

  1. Show that the variance of the random variable \(\hat{p}\) is \(\dfrac{7}{30\ 000}\).  (2 marks)

  2. Use the standard normal distribution and the normal distribution table to approximate \(P(\hat{p} \leq 0.31)\), giving your answer correct to two decimal places.  (2 marks)

Show Answers Only

  1. \(\text{See Worked Solutions}\)
  2. \(0.74\)

Show Worked Solution

i.    \(\text{Let}\ X =\ \text{number of people born overseas} \)

\(X\sim \text{Bin}(900,0.3) \)

\(\hat{p} = \dfrac{X}{900}\)

\(\text{Var}(X) = np(1-p)=900 \times 0.3 \times 0.7 \)

\(\text{Var}(\hat{p})\) \(=\dfrac{\text{Var}(X)}{n^2} \)  
  \(=\dfrac{900 \times 0.3 \times 0.7}{900^2}\)  
  \(=\dfrac{7}{30\ 000} \)  

 

ii.    \(\mu = E(\hat{p}) = 0.3,\ \ \sigma(\hat{p}) = \sqrt{\dfrac{7}{30\ 000}} \)

\(z\text{-score (0.31)} \ = \dfrac{0.31-0.3}{\sqrt{\dfrac{7}{30\ 000}}} = 0.6546… \)

\(P(\hat{p} \leq 0.31) \) \(=P(z \leq 0.6546…) \)  
  \(=0.74\ \ \text{(to 2 d.p.)} \)  

Statistics, EXT1 S1 2022 HSC 13e

A chocolate factory sells 150-gram chocolate bars. There has been a complaint that the bars actually weigh less than 150 grams, so a team of inspectors was sent to the factory to check. They randomly selected 16 bars, weighed them and noted that 8 bars weighed less than 150 grams.

The factory manager claims 80% of the chocolate bars produced by the factory weigh 150 grams or more.

  1. The inspectors used the normal approximation to the binomial distribution to calculate the probability, `cc "P"`, of having at least 8 bars weighing less than 150 grams in a random sample of 16, assuming the factory manager's claim is correct.
  2. Using the attached probability table, calculate the value of `cc "P"`.   (2 marks)

  3. The factory manager disagrees with the method used by the inspectors as described in part (i).
  4. Explain why the method used by the inspectors might not be valid.   (1 mark)

Show Answers Only
  1. `P=0.0013`
  2. `text{See Worked Solutions}`
Show Worked Solution

i.   `text{Let}\ \ X=\ text{number of chocolate bars < 150 grams}`

`X\ ~\ text{Bin}(16, 0.2)`

`mu=np=16xx0.2=3.2`

`sigma^2=np(1-p)=16xx0.2xx0.8=2.56`

`ztext{-score}\ (X=8)=(8-3.2)/sqrt2.56=3`

 

`cc P(X>=8)` `=P(z>=3.0)`  
  `=1-P(z<=3.0)`  
  `=1-0.9987`  
  `=0.0013`  

 

ii.   `np=3.2,\ \ n(1-p)=16xx0.8=12.8`

`text{S}text{ince}\ \ np<5,\ text{the sample size is too small.}`


♦ Mean mark part (i) 37%.
 
Mean mark part (ii) 54%.

Statistics, EXT1 S1 2021 HSC 14d

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)

Show Answers Only

`6000`

Show Worked Solution

`E(overset^p) = p = 3/500 = 0.006`

♦♦ Mean mark 27%.

`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`

Statistics, EXT1 S1 EQ-Bank 14

It is known that 65% of adults over the age of 60 have been tested for bowel cancer.

A random sample of 140 adults aged over 60 years is surveyed.

Using a normal approximation to the binomial distribution and the probability table attached, calculate the probability that at least 85 of the adults chosen have been tested for bowel cancer.   (3 marks)

Show Answers Only

`0.8554`

Show Worked Solution

`text(Let)\ \ X =\ text(number tested for cancer)`

`X\ ~\ text(Bin)(140, 0.65)`

`text(Find)\ \ P(X>=85):`

`E(hat p)` `=p=0.65`  
`text(Var)(hat p)` ` = (0.65(0.35))/140 = 0.001625`  
`sigma(hat p)` `=0.04031…`  

 
`hatp\ ~\ N(mu,sigma)\ ~\ N(0.65, 0.04031…)`

`text(If)\ \ X=85\ \ => \ hatp=85/140=0.60714…`

`ztext{-score}\ (X=85)` `=(0.60714-0.65)/0.04031`  
  `~~ -1.063`  

 
`text{Using the probability table (attached):}`

`P(z<–1.06) = 0.1446`

`:. P(X>=85)` `=1-0.1446`  
  `=0.8554`  

Statistics, EXT1 S1 EQ-Bank 22

It is known that 43% of voters in a city voted for Labor in the election.

If 250 voters from the city are selected at random, use a suitable approximation and the attached probability table to find the probability that at least 120 of those chosen voted for Labor.  (3 marks)

Show Answers Only

`0.0571`

Show Worked Solution

`E(overset^p) = p = 0.43`

`text(Var)(overset^p) = (0.43(1 – 0.43))/250 = 0.0009804`

`sigma(overset^p) = 0.0313`
 

`text(Let)\ \ X = text(number who voted Labor)`

`text(If)\ \ X = 120 \ => \ overset^p = 120/250 = 0.48`

`P(X >= 120)` `= P(z >= (0.48 – 0.43)/0.0313)`
  `= P(z >= 1.58)`
  `= 1 – 0.9429`
  `= 0.0571`

Statistics, EXT1 S1 EQ-Bank 20

Netball Australia records show that 10% of all registered players are over the age of 25.

  1. A random survey of 100 netball players was carried out to find out how many were over 25 years of age.

     

    Assuming the sample proportion is normally distributed, calculate the expected mean and standard deviation of this group.  (2 marks)

  2. Using the probability table attached, estimate the probability that at least 15 players surveyed will be over 25 years of age.  (2 marks)

Show Answers Only
  1. `0.03`
  2. `0.0475`
Show Worked Solution
i.   `E(hat p)` `= p = 0.1`
  `text(Var)(hat p)` `= (p(1 – p))/n`
    `= (0.1(0.9))/100`
    `= 0.0009`
  `sigma(hat p)` `= sqrt(0.0009)`
    `= 0.03`

 

ii.  `Ptext{(at least 15 players are over 25)} = P(hat p >= 0.15)`

`hat p\ ~\ N(mu,sigma)\ ~\ N(0.1, 0.03)`

`P(hat p >= 0.15)` `= P(z >= (0.15 – 0.10)/0.03)`
  `= P(z >= 1.67)`
  `= 1 – 0.9525`
  `= 0.0475`

Statistics, EXT1 S1 SM-Bank 11

Within a particular population, it is known that the percentage of left-handers is 17%.

A research project randomly selects 200 people from this population.

Assuming this sample proportion is normally distributed, what is the probability that the percentage of people that are left-handed in this sample is

  1. greater than 20%   (2 marks)

  2. less than 10%   (2 marks)

Show Answers Only
  1. `0.1292`
  2. `0.0041`
Show Worked Solution

i.   `E(hat p)=p=0.17`

`text(Var)(hat p)` ` = (p(1-p))/n`  
`sigma^2(hat p)` `=(0.17(1-0.17))/200`  
  `=0.0007055…`  
`sigma(hat p)` `=0.02656…`  
     

`hatp\ ~\ N(mu,sigma)\ ~\ N(0.17, 0.02656)`

`ztext{-score (20%)}` `=(x-mu)/sigma`  
  `~~(0.20 – 0.17)/0.02656`  
  `~~1.13`  

 
`text{Using the probability table (attached):}`

`:.P(hat p>20text(%))` `=P(z>1.13)`  
  `=1-0.8708`  
  `=0.1292`  

 

ii.    `ztext{-score (10%)}` `~~(0.10-0.17)/0.02656`
    `~~-2.64`

 
`text{Using the probability table (attached):}`

`:.P(hatp<10text(%))` `=P(z<-2.64)`  
  `=0.0041`  

Statistics, EXT1 S1 SM-Bank 6

After counting votes in an election, it is known that 40% of people voted for the Katter party.

A sample of 600 voting cards are taken and inspected. Assuming this sample proportion is approximately normally distributed, what is the probability that the percentage of voting cards inspected that chose the Katter party is less than 36%?   (3 marks)

Show Answers Only

`2.5\ text(%)`

Show Worked Solution

`mu = E(hat p)=p=0.4`

`text(Var)(hat p)` ` = (p(1-p))/n`  
`sigma^2(hat p)` `=(0.4(1-0.4))/600`  
  `=0.0004`  
`sigma(hat p)` `=0.02`  
     
`ztext{-score (36%)}` `=(0.36-0.40)/0.02`  
  `=-2`  

 

`:.P(hat p<36text(%))` `=P(z<-2)`  
  `=2.5 text(%)`