P(X=x) = e^(-mu) * mu^x / x! for x = 0, 1, 2, ...
In a laboratory experiment, there is a small chance of a bacterial colony appearing at a precise location on an agar plate. Furthermore, locations of different bacterial colonies are independent of each other. If you expect on average to see 0.02 bacterial colonies per cm^{2}, how likely is it to find 0, 1, 2, ... bacterial colonies on a 100 cm^{2} agar plate?
Solution: The number bacterial colonies may have a Poisson distribution.
The expected number, or mean is 100*(0.02) = 2.
P(X=0) = e^(-2) * 2^0 / 0! = 0.1353
P(X=1) = e^(-2) * 2^1 / 1! = 0.2707
P(X=2) = e^(-2) * 2^2 / 2! = 0.2707
P(X=3) = e^(-2) * 2^3 / 3! = 0.1804
and so on.
Solution: The mean is 2, the standard deviation is sqrt(2) = 1.414.
z = (x-mu)/sigma
.
In particular, the probability a normal random variable x
with mean mu and standard deviation sigma
is between numbers a
and b
is equal to the probability that a standard normal random variable z
is between (a-mu)/sigma
and (b-mu)/sigma
.
The normal table here does not have information for negative z values. Because of symmetry, this is not necessary.
Area to the left of a positive z.
P(Z < 2.05) = 0.9798.
Area to the left of a negative z.
P(Z < -2.05) = P(Z > 2.05) = 1 - 0.9798 = 0.0202.
Area between two positive z scores.
P(1.23 < Z < 2.05) = P(Z < 2.05) - P(Z < 1.23) = 0.9798 - 0.8907 = 0.0891.
Area between a positive and a negative z score.
P(-1.23 < Z < 2.05) = P(Z < 2.05) - P(Z < -1.23)
= P(Z < 2.05) - P(Z > 1.23)
= 0.9798 - (1 - 0.8907) = 0.8705.
Area outside two z scores.
P(|Z| > 1.23) = P(Z < -1.23) + P(Z > 1.23)
= 2*P(Z > 1.23)
= 2*(1 - 0.8907) = 0.2186.
The area to the left of z=0.52 is 0.6985 and the area to the left of z=0.53 is 0.7019.
The z-score we want is about half-way in between these, say z = 0.525.
Percentiles for a negative z.
Find the number z such that the area to the left of z is 0.1000.
This z will be negative. The corresponding positive z has an area to the right of 0.1000, or an area to the left of 0.9000.
The closest such z is z=1.28.
Therefore, the 10th percentile is at about z=-1.28.
Finding the cut-offs of a middle area.
Find the number z such that the area between -z and z is 0.7000.
The middle 70% leaves 30% left over, half on each side. Thus -z is at the 15th percentile and z is at the 85th percentile.
The closest such z is z=1.04.
Therefore, the area between -1.04 and 1.04 is about 70%.
To solve any problem for an arbitrary normal curve, translate it to a problem with the standard normal curve. Standardize with the formula
z = (x-mu)/sigma
or
x = mu + z*sigma
For these problems, assume that mu = 500 and sigma = 100.
Area to the left.
P(X < 350) = P(Z < (350-500)/100) = P(Z < -1.50) = P(Z > 1.50) = 1 - 0.9332 = 0.0668.
Area between two values.
P(450 < X < 720) = P((450-500)/100 < Z < (720-500)/100) = P(-0.50 < Z < 2.20)
= P(Z < 2.20) - P(Z > 0.50)
= 0.9861 - (1-0.6915) = 0.6776.
Area outside two values.
P(|X-500| > 161) = P(X < 339) + P(X > 661)
= P(Z < (339-500)/100) + P(Z > (661-500)/100)
= P(Z < -1.61) + P(Z > 1.61)
= 2*P(Z > 1.61)
= 2*(1-0.9463) = 0.1074.
Percentiles.
Find the value x that cuts off the top 40% of the values.
If x cuts off the top 40%, it cuts off the bottom 0.6000 area.
The z score is close to 0.25, so the x score is 0.25 standard dviations above the mean.
x = 500 + 0.25(100) = 525.
Middle area.
Find the values x and y that cut off the middle 40% of the values.
There is 60% left over, 30% on ech side. Thus, x is the 30th percentile and y is the 70th percentile.
The z-score of y is 0.525 and the z-score of x is -0.525.
x = 500 - 0.525(100) = 447.5 and y = 500 + 0.525(100) = 552.5.
Bret Larget, larget@mathcs.duq.edu