dp studies y2 chapter 10: normal distribution. contents: a. the normal distribution b. probabilities...
TRANSCRIPT
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.
A. The Normal Distribution
If a continuous variable X is normally distributed with meanm and standard deviation s, we write X ~ N(m, s2).
A. The Normal Distribution
A. The Normal Distribution
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.
A. The Normal Distribution
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.
B. Probability using a calculator
For the Ti-84:2nd VARS = DISTR
Choose “2” = normalcdf (“lowest bound”, “highest bound”, mean, standard deviation)
B. Probability using a calculator
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)
B. Probability using a calculator
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.
B. Probability using a calculator
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)
C. Quantiles or k-Values
Example 4:If X ~ N(23.6, 3.12), find k for which P(X < k)
= 0.95
C. Quantiles or k-Values
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
C. Quantiles or k-Values
To deal with P(X > k) = p, we use P(X < k) = 1 – p
C. Quantiles or k-Values
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.