Approximating Binomial Probabilities

Under certain sample size and proportion conditions, we can use the Normal Distribution (z-values) to approximate sample statistics and their probabilities from a discrete proportional distribution like the Binomial.

When the sample size n is sufficiently large and

if 5 and 5,

the distribution of a proportion is approximately Normal.

We use these formulas for the mean and standard deviation:

The Normal Distribution is continuous, so any probability value in it represents the AREA under the curve on the given interval. The Normal distribution, gives us the probability that
x is between 2 given values.

Should we need to find the probability that x is exactly equal to a given constant, say 7, we find the probability that x lies between 6.5 and 7.5.

The Binomial distribution is discrete, so to find the probability that the random variable takes on a specific value, we must assume that value extends over the interval from 0.5 below it to 0.5 above it. So, if we want the probability that 28, we find the probability that 27.5.

Continuity Correction:

In order to apply the properties of a continuous distribution to a discrete one,
we represent every whole number k by the interval k – ½ to k + ½.

To find P( x = 5), we find P ( 4.5 < x < 5.5).

To find P( x > 5), we find P ( x > 5.5).

To find P( x < 5), we find P( x < 4.5).

To find , we find .

We represent x = 1 with the interval

Example:

The Wayward Inn, a 300-room resort has experienced an 85% occupancy rate in the month of January for the past 10 years. The management is questionning whether this has changed so before they commit a lot of money to redecorating, they need to know:

a) the probability that at least 260 rooms will be occupied next January.

Since n = 300, and p = 0.85, we can see that 5 and 5,
we'll use the Normal Approximation of this Binomial situation.

Since we want , we'll set x = 259.5 and find .

Solution:

a)

Now we need P(z > 0.73) from the Normal Distribution table.

P(z > 0. 81) = 0.5000 – 0.2673 = 0.2327

The probability that at least 260 rooms will be occupied next January is 23.27%

b) the probability that fewer than 240 rooms will be occupied next January?

Solution:

Since we want , we'll set x = 239.5 and find

In this case, so we find z for this value.

Now we need P(z < – 2.51) from the Normal Distribution table.

P(z < – 2.51) = 0.5000 – 0.4940 = 0.0060

The probability that less than 240 rooms will be occupied next January is 0.6%.

Note: how we used 259.5 for 260 and 239.5 for 240 using the continuity correction factor.

Approximating Poisson Probabilities

We generally use the Poisson Distribution to count the number of successes over a fixed interval of time or within a specified region. For example: the number of murders recorded in Detroit in a month, or the number of cars parked at the airport during a 48 hour period.

The mean µ = np, as with the Binomial Distribution.

The standard deviation

The formula for calculating Poisson probabilities is:

where µ = the mean or expected number of successes,
e = the constant (2.718..), and x = the number of successes observed or counted.

Note: many statistics books use (lambda) instead of l for the expected mean value.

We can approximate the values of a Poisson Distribution
with the Normal Distribution when
n - the sample size is large, ...... p - the probability of a success is small,
and µ > 3, ...... with µ = np.

As with the Binomial Approximation, we use continuity correction.

Example: During the ski season, the medical clinic at Mont Tremblant Village treats an average of 3 patients with a broken bone per week, during the 8-hour day shift. If the number of patients per week with a broken bone approximates a Poisson distribution, find the probability that next week, the clinic will treat between 2 and 5 (inclusive) such patients. Use both the Poisson Distribution Table and the Normal Approximation, then compare the results.

Solution:

We want = P(2) + P(3) + P(4) + P(5)

Using the table: = 0.2240 + 0.2240 + 0.1680 + 0.1008 = 0.7168

Using the Normal Approximation with continuity correction:
we set x = 1.5 for 2 and x = 5.5 for 5 since 2 and 5 are included.

and

Since z ₁ is left of the mean and z ₂ is right of the mean, we add the probabilities.

P(z_{– 0.87} ) = 0.3078, and P(z _1.44 ) = 0.4251

0.3078 + 0.4251 = 0.7329, which is 0.0161 greater than 0.7168.

Practice

Use the Normal Approximation of the Binomial Distribution with Continuity Correction
for questions 1 to 6. Make a diagram, state µ and .

1) Records show that 25% of the cars manufactured at a certain plant need repairs in the
first 3 years of operation. If the IBM Corporation buys 90 of these cars,
find the probability that:

a) in the first 3 years of operation, fewer than 24 of the cars need repairs.

b) between 20 and 23 (inclusive) of the cars need repairs.

2) The yearly number of major earthquakes in the world is a random variable having approximately the normal distribution with µ = 20.8 and s = 4.5.

Find the probabilities that in any given year, there will be:

a) exactly 18 major earthquakes;

b) at least 18 major earthquakes;

c) at most 16 major earthquakes.

3) Research shows that a 3-foot high elm tree we transplant in the spring, has a 40% chance of surviving its first winter. If we transplant 50 such trees:

a) What is the probability that 25 or more of them survive their first winter?

b) What is the probability that between 18 and 23 of them survive their first winter?

4) Tests have shown that a new allergy drug is effective in 90% of the patients taking it. If the drug is administered to 80 allergy sufferers, what is the probability that it will be effective for at least 70 of them?

5) A certain contraceptive device is effective 90% of the time if used correctly.
If the device is used 300 times:

a) how many times should we expect it to fail?

b) what is the probability that it will fail 35 or more times?

6) Canada Post claims that 80% of the letters mailed in Montreal, destined for Vancouver will be delivered within 3 working days. If you mail 200 such letters, find the probability that:

a) more than 150 of them will arrive within 3 working days.

b) fewer than 148 of them will arrive within 3 working days.

c) between 150 and 160 of them will arrive within 3 working days.

d) What's the minimum number of letters that arrived on time if the probability of this event is 10% or more?

Use the Normal Approximation of the Poisson Distribution with Continuity Correction
for questions 7 and 8. Make a diagram, state µ and .

7) A parking lot attendant says he parks on average 12 cars per hour. What is the probability that he will park more than 15 cars between 3 and 4 p.m. today?

8) The Royal Bank guichet at the corner of Cavendish and Somerled is used 15 times per hour on an average day. Find the probability that more than 12 people will use this guichet between 4 and 5 p.m. today if today is considered an average day.

Solutions

1) µ = np = 22.5, and

a) for fewer than 24 cars, we use x = 23.5 so,

z _0.24 = 0.0948, so P(x < 24) = 0.5 + 0.0948 = 0.5948

_________________________

b) for between 20 and 23 cars inclusive, we use x = 19.5 for 20 and x = 23.5 for 23 (see part a)

so, .

so P(20 < x < 22.5) = z _0.73 = 0.2673, and P(22.5 < x < 23) = 0.0948 from part (a).

so P(20 < x < 23) = 0.0948 + 0.2673 = 0.3621

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2) µ = 20.8 and s = 4.5 (these were given)

a) for exactly 18, using Continuity Correction, we find P(17.5 < x < 18.5)

.

P(– 0.73 < z < 0) = 0.2673; and P(– 0.51 < z < 0) = 0.1950, therefore

P(17.5 < x < 18.5) = 0.2673 – 0.1950 = 0.0723

_________________________

b) for P(x 18), we set x = 17.5 and in (a) we got P(17.5 < x < 20.8) = 0.2673
When we add 0.5000 for P(x > 20.8), we get P(x 18) = 0.7673.

_________________________

c) for at most 16, we find P(x < 16.5)

and P(– 0.96 < z < 0) = 0.3315

P(x < 16.5)= 0.5000 – 0.3315 = 0.1685.

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3) µ = 50(.40) = 20, and

a) for , we use x = 24.5

z = (24.5 – 20)/3.464 = 1.30, and P(z > 1.30) = 0.5000 – 0.4032 = 0.0968

____________________________

b) P(18 < x < 23) becomes P(17.5 < x < 23.5) with continuity correction.

z = (17.5 – 20)/3.464 = – 0.72 ....... and ....... z = (23.5 – 20)/3.464 = 1.01

P(– 0.72 < z < 1.01) = 0.2642 + 0.3438 = 0.6080

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4) µ = 80(.90) = 72, and

for , we use x = 69.5

z = (69.5 – 72)/2.6833 = – 0.93, and P(z > – 0.93) = 0.5000 + 0.3238 = 0.8238

____________________________

5) µ = 300(.10) = 30, and

a) P(effective) = 0.90, so P( failure) = 0.10, so we expect it to fail 30 times in 300.

____________________________

b) for , we use x = 34.5

z = (34.5 – 30)/5.1962 = 0.87, and P(z > 0.87) = 0.5000 – 0.3078 = 0.1922

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6) µ = 200(.80) = 160, and

a) for , we use x = 149.5

z = (149.5 – 160)/5.6569 = – 1.68, and P(z > – 1.68) = 0.5000 + 0.4535 = 0.9535

____________________________

b) for , we use x = 147.5

z = (147.5 – 160)/5.6569 = – 2.21, and P(z < – 2.21) = 0.5000 – 0.4864 = 0.0136

____________________________

c) for , we use x = 149.5 and x = 160.5 respectively

z₁ = (149.5 – 160)/5.6569 = – 1.86, and P(– 1.86 < z < 0) = 0.4686

z₂ = (160.5 – 160)/5.6569 = 0.09, and P(0 < z < 0.09) = 0.0359

the probability that = 0.4686 + 0.0359 = 0.5045

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7) µ = 12 (given), and

for , we use x = 15.5

z = (15.5 – 12)/3.4641 = 1.01, and P(z > 1.01) = 0.5000 – 0.3438 = 0.1562

_________________________________________

8) µ = 15 (given), and

for , we use x = 12.5

z = (12.5 – 15)/3.8730 = – 0.65, and P(z > – 0.65) = 0.5000 + 0.2422 = 0.7422

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