Since we want to know the probability of more than 45 people winning, the boundary is 44.5. Using this, we can find the z-score using the following equation: Z(44.5) = Z(44.5) = = 1.24Using the normal table, we can see that the probability of the outcome having a z-score less than 1.24 is 0.8925. We can subtract this number from 1 to find the probability of there being morethan 45 wins. 1-1-1 - 0.8925 = 0.1075 Therefore, there is only a 10.75% chance that Yuen Zhi will not have enough prizes. She will probably be okay. 9. a) We can find probability that a flight is delayed 20 times in the month of June, when there is a 0.3 probability that it is delayed using both binomial and normal. Using normal, the solution would be as follows: n= 30 and p= 0.6μ = np= 30 × 0.6 = 18σ = √np(p− 1) = √20 × 0.6 × 0.7 = 2.68The boundaries are 19.5 and 20.5Z(19.5) = = 0.56 ( probability is 0.7123)Z(20.5) = = 0.93 (probability is 0.8238) 0.8238 – 0.7123 = 0.1115 ∴The probability of the flight being delayed 20 times using the normal is 11.15% Using binomial, the solution would be as follows: p(20) = 30C200.620 0.410= 0.11519∴The probability of the flight being delayed 20 times using the binomial is 11.1519% From these two solutions we can see that both ways got the same solution rounded to 2 decimal places (11.15%) . However, using the binomial was more accurate as it gave a probability to 4 decimal places instead of 2. b) The normal would give a less accurate approximation if np ≤5 and n(1 – p ) ≤ 5c) If the criteria are met, it is better to use the binomial if the normal requires more work. This would be the case in situations where we are finding the probability of one single value. The situation explained in
part (a) can be an example of this, since we were only finding the probability of the flight being delayed 20 times. Since we can find a solution to this using only one calculation with the binomial, it is more efficient.
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