Download Formula sheet for Statistic test 1 and more Exercises Mathematics in PDF only on Docsity!
MATH 3339 Statistics for the Sciences
Test 1 Fomula Sheet
Descriptive Statistics
P
1 n^ x 1 , sample mean n i=
middle value of ordered data if n is odd
mean of two middle values of ordered data if n is even
P
2 1 n
2 , sample variance n− 1 i=
2 , sample standard deviation
- IQR = Q 3 − Q 1 , interquartile range where Q 3 is the 75
th percentile and Q 1 is the 25
th
percentile.
x−x¯
, standard score
• CV =
x
s
¯
, coeffcient of variation
P
1 n^ (x 1 − x¯)(yi − y¯), sample covariance n− 1 i=
cov(x,y)
, sample correlation
• R
2 = r
2 , coeffcient of determination
1 =^ r^ s
s
x
y , sample slope of least-squares regression line
0 =^ y¯^ −^ β
1 x¯, sample y-intercept of least-squares regression line
th observation
Counting and Probability
- n! = n · (n − 1) · (n − 2) · · · 1 , factorial
- P (n, r) =
n! , permutation (n−r)!
n! , combination r!(n−r)!
- P (A ∪ B) = P (A) + P (B) − P (A ∩ B), general addition rule for probability
P (A∩B)
- P (A|B) = , conditional probability P (B)
- Given that events Ai, i = 1, 2 ,... , n, are disjoint and exhaustive of the sample space S, then P
P (B) = i
n
=
P (B|Ai)P (Ai), law of total probability.
Discrete Probabilities
P
n
xipi, expected value of the distribution
P
2
n (xi − μx)
2 pi = E(X
2 ) − E(X)
2 x , variance of the distribution i=
p
2 , standard deviation of the distribution
- Binomial Distribution, n = number of trials, p = probability of success
k (1 − p)
n−k , probability of x successes in n independent trials
- E(X) = np, expected value of binomial distribution
- V ar(X) = np(p − 1), variance of binomial distribution
- Geometric Distribution, p = probability of success
- P (X = x) = (1 − p)
( x − 1)p, probability that the x
th trial is the frst success
x , probability that the frst success is more than the x
th trial
– E(X) =
1
p
expected value of geometric distribution
1
p
−p 2 , variance of geometric distribution
- Hypergeometric Distribution, m = number of successes in the population, n = number of
failures in the population, k = number of trails
C(m,x)×C(n,k−x)
- P (X = x) = , probability of x successes C(m+n,k)
– E(X) =
km , expected value of hypergeometric distribution m+n �
k− 1 , variance of hypergeometric distribution, where m=n− 1
k p = m+n
- Poisson Distribution, μ = mean number of successes in a unit
e
−
x
μ
!
μ
x
probability of x successes
- E(X) = μ, expected value for Poisson distribution
- V ar(X) = μ, variance for Poisson distribution
