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 cm2, how likely is it to find 0, 1, 2, ... bacterial colonies on a 100 cm2 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
with mean mu and standard deviation sigma
is between numbers
is equal to the probability that a standard normal random variable
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
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.
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.
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, firstname.lastname@example.org