DP Studies Y2 Chapter 10: Normal Distribution. Contents: A. The normal distribution B. Probabilities...

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DP Studies Y2

Chapter 10:

Normal Distribution

Contents:

A. The normal distributionB. Probabilities using a calculatorC. Quantiles or k-values

Opening Problem

A continuous random variable is a variable which can take any real value within a certain range. We usually denote random variables by a capital letter such as X. Individual measurements of this variable are denoted by the corresponding lower case letter x.

For example, the probability that an egg will weigh exactly 72.9 g is zero.

If you were to weigh an egg on scales that weigh to the nearest 0.1 g, a reading of 72.9 g means the weight lies somewhere between 72.85 g and 72.95 g. No matter how accurate your scales are, you can only ever know the weight of an egg within a range.

(think of a number line where you have all real numbers)

So, for a continuous variable we can only talk about the probability that an event lies in an interval, and:

P(a < X < b) = P(a < X < b) = P(a < X <b) = P(a < X < b).

A. The Normal Distribution

Although all normal distributions have the same general bell-shaped curve, the exact location and shape of the curve is determined by the mean m and standard deviation s of the variable. (Notice that the normal curve is always symmetric about the vertical line x = m.)

examples:

1. The height of trees in a park is normally distributed

with mean 10 meters and standard deviation

3 meters.

2. The time it takes Sean to get to school is

normally distributed with mean 15 minutes and

standard deviation 1 minute.

If a continuous variable X is normally distributed with meanm and standard deviation s, we write X ~ N(m, s2).

Example 1

The chest measurements of 18 year old male footballers are normally distributed with a mean of 95 cm and a standard deviation of 8 cm.

a Find the percentage of footballers with chest measurements between:

i. 87 cm and 103 cm

ii. 103 cm and 111 cm

b Find the probability that the chest measurement of a randomly chosen footballer is between 87 cm and 111 cm.

Solutions to example 1:

B. Probability using a calculator

Example 2:If X ~ N(10, 2.32), find these probabilities:

a. P(8 < X < 11) b. P(X < 12) c. P(X > 9). Illustrate your results.

For the Ti-84:2nd VARS = DISTR

Choose “2” = normalcdf (“lowest bound”, “highest bound”, mean, standard deviation)

Solutions to example 2:a. normalcaf(8, 11, 10, 2.3) = 0.4759

P(8 < X < 11) = 0.4759

b. normalcaf(-1e99, 12, 10, 2.3) = 0.8077P(8 < X < 11) = 0.4759

c. normalcaf(9, 1e99, 10, 2.3) = 0.6681P( X > 11) = 0.6681

Note: for continuous distributionP(X > 9) = P(X < 9)

Example 3:In 1972 the heights of rugby players were approximately normally distributed with mean 179 cm and standard deviation 7 cm. Find the probability that a randomly selected player in 1972 was:

a. at least 175 cm tall

b. between 170 cm and 190 cm.

Solutions to example 3:If X is the height of a player then X is normal distributed with m = 179, s = 7.a. b.

C. Quantiles or k-ValuesTo understand the definition of a quantile, we need to look at an example:

Consider a population of crabs where the length of a shell, X mm,is normally distributed with mean 70 mm and standard deviation10 mm.A biologist wants to protect the population by allowing only thelargest 5% of crabs to be harvested. He therefore asks the

question:“95% of the crabs have lengths less than what?”.

To answer this question we need to find k such thatP(X < k) = 0.95 .

The number k is known as a quantile, and in this case the 95% quantile.

C. Quantiles or k-Values

When finding quantiles we are given a probability and are asked to calculate the corresponding measurement. This is the inverse of finding probabilities, and we use the inverse normal function on our calculator.

invNorm(quantile, mean, standard deviation)

Example 4:If X ~ N(23.6, 3.12), find k for which P(X < k)

= 0.95

Solution to example 4:m = 23.6 and s = 3.1invNorm(0.95, 23.6, 3.1) = 28.69904624

Therefore P(X < 28.7) = 0.95

To deal with P(X > k) = p, we use P(X < k) = 1 – p

Example 5:A university professor determines that 80% of this year’s History candidates should pass the final examination. The examination results were approximately normally distributed with mean 62 and standard deviation 12. Find the lowest score necessary to pass the examination.

Solution to example 5:

invNorm(0.2, 62, 12) = 51.9005452k ≈ 51.9

So the minimum pass mark is 52.

DP Studies Y2 Chapter 10: Normal Distribution. Contents: A. The normal distribution B. Probabilities...

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