Using Normal Distributionto Approximate aDiscrete Distribution
Discrete Distributions
Probability Distributionof a discrete variable isrepresented by ahistogram
Discrete Distributions
In some cases
It can be approximated
By a normal distribution
The number of express mail packages mailed at a given office isapproximately normal with a mean of 18 and standard deviation of 6.
A Distribution hasApproximatelyNormal Distribution
6 12 18 24 30
What is the probability that x=20?
In a Normal Distribution, probability is represented by the area under acurve. So the probability that it is “exactly” 20 is zero.
The number of express mail packages mailed at a given office isapproximately normal with a mean of 18 and standard deviation of 6.
A Distribution hasApproximatelyNormal Distribution
6 12 18 24 30
What is the probability that x=20?
In a Normal Distribution, probability is represented by the area under acurve. The probability that it is between 19.5 and 20.5 is .0641, or 6.41%
Replace the whole number r
With the range r - 0.5 to r +0.5
This is the continuity correction.
Find the probability between
20- 0.5 and 20 + 0.5.
70 85 100 115 130
An IQ Score must be a whole number, so is representedby a discrete distribution.
Assume that the distribution can be approximated witha normal distribution with a mean of 100 and standarddeviation of 15.
IQ Scores
Find P(x=100)
(x=100) =>
(99.5 ≤ x ≤ 100.5)
70 85 100 115 130
An IQ Score must be a whole number, so is representedby a discrete distribution.
Assume that the distribution can be approximated witha normal distribution with a mean of 100 and standarddeviation of 15.
IQ Scores
Find P(x ≤ 110)
(x ≤ 110) =>
(x ≤ 110.5)
Find P(x < 110)
(x < 110) =>
(x ≤ 109)
(x ≤ 109) =>
(x ≤ 109.5)
70 85 100 115 130
An IQ Score must be a whole number, so is representedby a discrete distribution.
Assume that the distribution can be approximated witha normal distribution with a mean of 100 and standarddeviation of 15.
IQ Scores
Find P(75 ≤ x ≤ 125)
(75 ≤ x ≤ 125) =>
(74.5 ≤ x ≤ 125.5)
Premature Babies:
Premature babies are born more than 3 weeks early.
10% of live births in US are premature and we have a sample of 250births.
Normal Approximationof Binomial Distribution
The distribution can be approximated by a Normal Distribution,
because np = 250(.10) = 25 and nq = 250(.9) = 225 are both > 5.
Mean = np = 25
This is a binomial distribution because a baby either is premature or is not.
Premature Babies:
Premature babies are born more than 3 weeks early.
If 10% of live births in US are premature and we have a sample of250 births:
Normal Approximationof Binomial Distribution
The distribution can be approximated by a Normal Distribution,
With mean = 25 and standard deviation = 4.743
What is the probability that the number of premature births
in our sample will be between 15 and 30 (inclusive)?
Find P(15 ≤ x ≤ 30)
(15 ≤ x ≤ 30) =>
(14.5 ≤ x ≤ 30.5)
Premature Babies:
Normal Approximationof Binomial Distribution
The distribution can be approximated by a Normal Distribution,
With mean = 25 and standard deviation = 4.743
What is the probability that the number of premature births
in our sample will be between 15 and 30 (inclusive)?
Find P(15 ≤ x ≤ 30)
(15 ≤ x ≤ 30) =>
(14.5 ≤ x ≤ 30.5)
(-2.21 ≤ z ≤ 1.16)