NORMAL DISTRIBUTION. Normal distribution- continuous distribution. Normal density: bell shaped, unimodal- single peak at the center, symmetric. Completely described by its center of symmetry - mean μ and spread - standard deviation σ. μ.
Normal distribution- continuous distribution.
Random variable with normal distribution – normal random variable with mean μ and st. dev. σ: X~N(μ, σ)
Standard normal random variable: mean 0 and st. dev. 1: Z~N(0, 1)
CHANGING SCALE σ
CHANGING LOCATION μ
1=σ1 < σ2 = 2
0=μ1 < μ2 =1
Changes in mean/location cause shifts in the density curve along the x-axis.
Changes in spread/standard deviation cause changes in the shape of the density curve.
Areas under the normal curve are probabilities (as under any density curve).
Special areas under the normal density curve:
The range "within one/two/three standard deviation(s) of the mean" is highlighted in green.
The area under the curve over this range is the rel. frequency of observations in the range.
That is, 0.68/95/99.7 = 68%/95%/99.7% of the observations fall within one/two/three standard deviation(s) of the mean, or, 68%/95%/99.7% of the observations are between (μ – 1/2/3σ) and (μ + 1/2/3σ).
Normal probabilities = areas under the normal curve are tabulated for the standard normal distribution (front cover of your text book).
In looking for probabilities keep in mind:
Symmetry of the normal curve and P(Z=a)=0 for any a.
P(Z < 0.01) = 0.504
P(Z ≤ -0.01 ) = 0.496
P(Z < 0) = 0.5
P( Z < 2.92)= 0.9982
P(Z>2.92)=1-0.9982=0.0018 or, by symmetry =P(Z< - 2.92)=0.0018
P(-1.32< Z <1.2)=0.8849 – 0.0934=0.7915
SUMMARY OF RULES we used above: P(Z>a)=1-P(Z< -a)
P(a < Z < b) = P(Z < b)- P( Z < a)
IF X~N(μ, σ) then Z = ~N(0, 1) standard normal.
Example. Suppose that the weight of people in NV follows normal distribution with mean 150 and standard deviation 20 lb. Find the probability that a randomly selected Nevadan weighs
at most 160 lb; b) over 160 lb.
Solution. Let X= weight of a randomly selected Nevadan. X~N(150, 20).
Given that P(Z < p)=0.95 find p. Here p is called 95thpercentile of Z.
Inside the table I looked for 0.95.
Found 0.9495 and 0.9505.
Used z-value corresponding to
the midpoint (0.95) between the
two available probabilities 1.645.
If an available probability is closer
to the one we need, use the
z-value corresponding to
EXAMPLE. Suppose scores X on a test follow a normal distribution with mean 430 and standard deviation 100. Find 90th percentile of the scores, that is find score x such that P(X ≤ x)=0.9.
Solution. Since we start with a normal but NOT STANDARD normal distribution, we have to standardize at some point:
0.9 = P(X ≤ x) =
x - 430 =128
x = 558
90% of students scored 558 or less.
Height of women follows normal distribution with mean 64.5 and standard deviation of 2.5 inches. Find
a) The probability that a woman is shorter than 70 in.
b) The probability that a woman is between 60 and 70 in tall.
c) What is the height 10% of women are shorter than, i.e. what is the 10th percentile of women heights?
SOLUTION. X= women height; X~N(64.5, 2.5).
a) P(X <70)=P(Z< (70-64.5)/2.5)=P(Z<2.2)=0.9861
b) P( 60 < X < 70) = P( (60-64.5)/2.5) < Z < (70-64.5)/2.5)=P(-1.8< Z < 2.2)= P( Z <-2.2) – P( Z < -1.8) = 0.9861 – 0.0359 = 0.9502.
c) 10th percentile of X =?
0.1=P( X< x) = P( Z< (x-65.5)/2.5), so -2.33=(x-65.5)/2.5; x=59.675.
10% of women are shorter than 59.675 in.