The Normal Distribution

Introduction to ND (Gaussian Distribution)

• Blue and Red – Same mean, Different Std. Deviations• Blue and Green – Different mean, Different Std. Deviations• Blue and Black – Different means, Same Std. Deviations

Properties of ND• A ND curve is bell-shaped• The mean, median and mode are equal and are

located at the centre of distribution.• The curve is symmetric about the mean.• The curve is continuous• The curve never touches the x-axis• The total area under a ND curve is 1.00 (100%)

Standard Normal Distribution

• z is the "z-score" (Standard Score)• x is the value to be standardized• μ is the mean• σ is the standard deviation

• Z ~ N (0,1) is used to standardize the values of the normal distribution X ~ N (μ, )

Why standardize?Example: Professor Willoughby is marking a test.

Here are the students results (out of 60 points):

20, 15, 26, 32, 18, 28, 35, 14, 26, 22, 17

Most students didn't even get 30 out of 60, and most will fail.

The test must have been really hard, so the Prof decides to Standardize

all the scores and only fail people 1 standard deviation below the mean.

The Mean is 23, and the Standard Deviation is 6.6, and these are the

Standard Scores:

-0.45, -1.21, 0.45, 1.36, -0.76, 0.76, 1.82, -1.36, 0.45, -0.15, -0.91

Only 2 students will fail (the ones who scored 15 and 14 on the test)

Let’s Answer Some Questions• The random variable X normally distributed and is such that the

mean μ is three times the standard deviation σ. It is given that P (X < 25) = 0.648I. Find the values of μ and σII. Find the probability that, from 6random values of X, exactly 4 are

greater than 25.z = 0.38 = 0.38μ = 22.2, σ = 7.40

P (4) = 6C4 = 0.0967

• The mean of a certain normally distributed variable is four times the standard deviation. The probability that a randomly chosen value is greater than at least 5 is 0.15.

I. Find the mean and standard deviationII. 200 values of the variable are chosen at random. Find the probability that at

least 160 of these values are less than 5.z = 1.036 or 1.0371.036 = s = 0.993μ= 3.97

p = 0.85μ = 200x0.85=170,var = 200x0.85x0.15=25.5P (at least 160) = P ( z > = P(z > -2.079)= 0.981

The Normal Approximate to the Binomial Distribution• Conditions

• If X~B(n,p) and np>5, nq>5. then X’~N(np,npq)

• Continuity correction (CC)• Binomial (discrete) Normal (continuous)

Eg: X is the no. of heads in 12 tosses . X~B(12,0.5)np = 6(>5) X’~N(6,3)nq = 6(>5)

np>5 and nq>5

Example question.• In Scotland, in November on average 80% of days are cloudy.

Assume that the weather on any one day is independent of the weather on other days.

IRP May/June 2011 (Paper 62)

i) Use a normal approximation to find the probability of there being fewer than 25 cloudy days in Scotland in November (30 days)

np = 24, nqp = 4.8z = = 0.228prob: 0.590

ii) Give a reason why the use of a normal approximation is justified.

np and nq both >5