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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.)`