Cambridge International AS & A Level Mathematics: Probability & Statistics 2
P(19 , X ø 21) 20
24 20 ! e –24 2421 + 21 ! = 0.134
=
e
–24
Normal approximation
Comments
19.5 − 24 21.5 − 24 P(19.5 ø Y ø 21.5) = P øZø 24 24 = Φ( − 0.510) − Φ( −0.919)
19 is not included but 21 is included, so the range of values is 19.5 to 21.5.
= (1 – 0.6950) – (1 – 0.8209) = 0.126
E
Poisson
From these calculations we can see that the probabilities using the normal distribution as an approximation to the Poisson distribution give the same set of values, correct to 2 decimal places, as the calculations using the Poisson probability formula.
PL
WORKED EXAMPLE 2.6
Over the years, a biologist notes that a species of turtle lays on average 60 eggs in each nest. The number of eggs laid in each nest follows a Poisson distribution, and is independent of the number of eggs laid in other nests. Calculate the probability that in a randomly chosen nest there are: a exactly 50 eggs
b over 74 eggs
Answer Let X be the random variable ‘number of eggs in a nest’. Then X ~ Po(60) ≈ N(60, 60). 42
SA M
49.5 − 60 50.5 − 60 a P( X = 50) ≈ P øZø 60 6 0 = Φ( − 1.226) − Φ( −1.356)
c 40 eggs or less.
Always state the distribution you are using and any approximating distribution.
You could just use Poisson here; the question is included to highlight the process for a single value.
= (1 – 0.8899) – (1 – 0.9125) = 0.0226
74.5 − 60 b P( X . 74) = 1 – P( X , 75) ≈ 1 – P Z ø 60 = 1 − Φ(1.872) = 1 – 0.9693 = 0.0307
You could have written 1 – P( X ø 74). 74.5 is the value you need to work with from writing either , 75 or ø 74 .
40.5 − 60 c P( X ø 40) ≈ P Z ø 60 = Φ( −2.517) = 1 – 0.9941 = 0.0059
EXERCISE 2D
1 X ~ Po(42). Use the normal approximation to find: a P( X , 50)
b P( X ø 40)
c P( X = 45)
2 X ~ Po(50). Use the normal approximation to find: a P(52 , X , 56)
b P(50 ø X ø 52)
Original material © Cambridge University Press 2017