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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)

  3. By using a normal approximation, find the approximate probability that `X` is between 55 and 65.  (1 mark)

Show Answers Only
  1. `60`
  2. `text(See Worked Solutions)`
  3. `68text(%)`
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`

 

iii.    `P(55 <= x <=65)` `~~ P(−1 <= z <= 1)`
    `~~ 68text(%)`

Statistics, EXT1 S1 2019 MET1 6

Jacinta tosses a coin five times.

  1. Assuming that the coin is fair and given that Jacinta observes a head on the first two tosses, find the probability that she observes a total of either four or five heads.  (2 marks)

  2. Albin suspects that a coin is not actually a fair coin and he tosses it 18 times.

     

    Albin observes a total of 12 heads from the 18 tosses.

  3. Let  `X` = probability of obtaining a head.
  4. Find the range of `X` in which 95% of observations are expected to lie within.  (2 marks)

Show Answers Only
  1.  `1/2`
  2.  `(4/9, 8/9)`
Show Worked Solution

a.   `text(After 2 tosses, 2 heads.)`

`Ptext{(4 or 5}\ H)` `= HHT + HTH + THH + HHH`
  `= (1/2)^3 xx 4`
  `= 1/2`

 

b.    `E(hat p) = p=12/18 = 2/3`

`text(Var)(hatp) = (2/3(1-2/3))/18=1/81`

`sigma(hatp)=1/9`

`text(95% interval)\  (z= +-2)`

`=(hatp-z xx sigma(hatp), \ hatp+z xx sigma(hatp))`

`= (2/3 – 2 xx 1/9, 2/3 + 2 xx 1/9)`

`= (4/9, 8/9)`

Statistics, EXT1 S1 EQ-Bank 23

A light manufacturer knows that 6% of the light bulbs it produces are defective.

Light bulbs are supplied in boxes of 20 bulbs. Boxes are supplied in pallets of 120 boxes.

Calculate the probability that

  1. A box of light bulbs contains exactly 3 defective bulbs.  (1 mark)

  2. A box of light bulbs contains at least 1 defective bulb.  (1 mark)

  3. A pallet contains between 90 and 95 (inclusive) boxes with at least 1 defective bulb (use the probability table attached).  (3 marks)

Show Answers Only
  1. `0.086\ \ (text(to 3 d.p.))`
  2. `0.710\ \ (text(3 d.p.))`
  3. `0.1416`
Show Worked Solution

i.   `P(D) = 0.06, \ P(barD) = 0.94`

`P(D = 3)` `= \ ^20C_3(0.06)^3(0.94)^17`
  `= 0.086\ \ (text(to 3 d.p.))`

 

ii.    `P(D >= 1)` `= 1 – P(D = 0)`
    `= 1 – \ ^20C_0(0.06)^0(0.94)^20`
    `= 0.710\ \ (text(to 3 d.p.))`

 

iii.   `text(Let)\ \ X = text(number of boxes where)\ \ D >= 1`

`text(Let)\ \ overset^p = text(proportion of boxes where)\ \ D >= 1`

`E(overset^p) = p = 0.710`

`text(Var)(overset^p) = (0.710(1 – 0.710))/120 = 0.0017158`

`sigma(overset^p) = 0.04142`
 

`overset^p\ ~\ N(0.710, 0.04142)`

`text(If)\ \ X = 90 \ => \ overset^p = 90/120 = 0.75`

`text(If)\ \ X = 95 \ => \ overset^p = 95/120 = 0.79167`
 

`ztext(-score)\ (X = 90) = (0.75 – 0.710)/(0.04142) = 0.9657`

`ztext(-score)\ (X = 95) = (0.79167 – 0.710)/(0.04142) = 1.972`
 

`P(90 <= X <= 95)` `= P(0.97 <= z <= 1.97)`
  `= 0.9756 – 0.8340`
  `= 0.1416`

Statistics, EXT1 S1 EQ-Bank 21

A biased coin has a 0.6 chance of landing on heads. The coin is tossed 15 times.

  1. Calculate the probability of obtaining 7, 8 or 9 heads using binomial probability distribution.
  2. Give your answer correct to 3 decimal places.  (2 marks)

  3. Calculate the probability of obtaining 7, 8 or 9 heads using normal approximation to the binomial distribution and the probability table attached.
  4. Give your answer correct to 3 decimal places.  (2 marks)

  5. Could this binomial distribution be reasonably approximated with a normal distribution? Support your answer with a brief calculation.  (1 mark)

Show Answers Only
  1. `0.502\ (text(to 3 d.p.))`
  2. `0.509`
  3. `text(See Worked Solutions)`
Show Worked Solution
i.    `P(7, 8\ text(or)\ 9)` `= \ ^15C_7(0.6)^7(0.4)^8  \ ^15C_8(0.6)^8(0.4)^7 + \ ^15C_9(0.6)^9(0.4)^6`
    `= 0.118056 + 0.177084 + 0.206600`
    `= 0.502\ \ (text(to 3 d.p.))`

 

ii.   `E(overset^p) = p = 0.6`

`text(Var)(overset^p) = (p(1 – p))/n = (0.6 xx 0.4)/15 = 0.016`

`sigma (overset^p) = 0.12649`

 
`text(Let)\ \ X = text(number of heads)`

`P(7, 8, 9) = P(6.5 < X < 9.5)`

`text(If)\ \ X = 6.5 \ => \ overset^p = 6.5/15 = 0.4333`

`text(If)\ \ X = 9.5 \ => \ overset^p = 9.5/15 = 0.6333`

`ztext(-score)\ (X = 6.5) = (0.4333 – 0.6)/0.12649 = −1.32`

`ztext(-score)\ (X = 9.5) = (0.6333 – 0.6)/0.12649 = 0.26`

COMMENT: A quick sketch of the normal distribution curve is often helpful when using probability tables in this context.
 

`P(7, 8, 9)` `= P(−1.32 < z < 0.26)`
  `= P(z < 1.32) – P(z < −0.26)`
  `= 0.9066 – 0.3974`
  `= 0.509`

 

iii.   `np = 15 xx 0.6 = 9`

COMMENT: `np>10, nq>10` is also common in assessing if normal approximation is reasonable.

`nq=n(1 – p) = 15 xx 0.4 = 6`

`text(S)text(ince)\ \ np > 5 and nq > 5,`

`=>\ text(Normal approximation is reasonable.)`

Statistics, EXT1 S1 SM-Bank 8

A laptop's battery is considered faulty if its battery life is less than 3 hours.

The laptop supplier knows that the chance of a faulty battery in any laptop is 6.5%.

A random sample of 70 laptops is selected from the supplier and the battery life of each laptop is tested.

Assuming the sample proportion is normally distributed, what is the probability that the percentage of laptops with faulty batteries lies between 5% and 10%?

Give your answer to the nearest percentage.   (3 marks)

Show Answers Only

`58 text(%)`

Show Worked Solution

`E(hat p)=p=0.065`

`text(Var)(hat p)` ` = (p(1-p))/n`  
`sigma^2(hat p)` `=(0.065(1-0.065))/70`  
  `=0.0008682…`  
`sigma(hat p)` `=0.029465…`  
     

`hatp\ ~\ N(mu, sigma)\ ~\ N(0.065, 0.029465)`

`ztext{-score (5%)}` `~~(0.05 – 0.065)/0.029465`  
  `~~-0.509`  
     
`ztext{-score (10%)}` `~~(0.10 – 0.065)/0.029465`  
  `~~1.187`  

 
`text{Using probability tables (attached)}:`

`:.P(5 text(%)<=hat p<=10text(%))` `~~P(-0.51<=z<=1.19)`  
  `~~0.8830-0.3050`  
  `~~0.5780`  
  `~~58text(%)`