STAT 263 term test to ch 9
SK =
3(mean − median)
standard deviation
V 1.7
index p th percentile =
z=
IQR = Q − Q
3
1
x−µ
σ
or
Revised 10/15/2008
p
( n + 1)
100
z=
x−x
s
s
σ
CV = ⋅100%
CV = ⋅100%
or
Coefficient of Variation:
x
µ
1
Chebyshev's Rule: at least 1 − 2 of the data fall within k standard deviations of the mean
k
x ≈ L +
Median for grouped data (similar for any fractile):
j
c
f
Where L is the lower boundary of the class into which the median must fall, f is the frequency of this
class, c is the class interval, and j is the number of values we still lack when we reach L.
P ( A) ≥ 0 for any event A; P ( A) ≤ 1 for any event A;
P (φ ) = 0;
P ( A) + P ( A′) = 1
If A and B are mutually exclusive events, then P(A ∪B)=P(A)+P(B).
If k events are mutually exclusive, then
P(A1 ∪ A2 ∪ A3 ∪ …. ∪ Ak )=P(A1)+P(A2) +…+ P(Ak)
It is always true that: P(A ∪B)=P(A)+P(B) - P(A ∩ B)
If P(B) is not equal to zero, then:
P( A B) =
P( A ∩ B)
P( B)
If A and B are independent: P(A) = P(A|B) and P(A ∩ B) = P(A) P(B)
It is always true that: P ( A ∩ B ) = P ( A B ) P ( B ) = P ( B A ) P ( A )
Bayes’ Theorem:
P ( A B) =
P (B A) P (A)
P ( B A ) P ( A ) + P ( B A′ ) P ( A′ )
P(at least one success) = 1 – P(zero successes)
If the probabilities of obtaining the amounts a1, a2, a3, … , or ak are p1, p2, p3, … , and pk , where
p1 + p2 + p3 + … + pk = 1, then the mathematical expectation is E = a p + a p + …+ a p
1 1
2 2
k k
x
n – x for x = 0, 1, 2, …, or n, µ = np, σ2 =np(1 − p)
Binomial: f(x) = nCx p (1 – p)
e−λ λ x
P ( X = x) = f ( x) =
, µ =λ
Poisson:
x!
a C x b Cn − x
Hypergeometric: P( X = x) = f ( x) =
for x = 0, 1, 2, …, or n; x < a, (n – x) < b
a + b Cn
Poisson Approx. to the Binomial
e − np (np ) x
P( X = x) ≈ f ( x) =
assumes np < 10, n > 100
x!