Stats 6.2 Vocab

variance

= np(1-p)

fewer than 3 wireless only

x=0,1,2 P(0) + P(1) + P(2) binompdf (20,.41, 0) + binompdf(20,.41, 1) +binompdf (20,.41,2) OR binomcdf P(<3)= binomcdf (20,.41,2)

Must satisfy 4 properties:

1. fixed number of trials n = 2. trials are independent 3. for each trial there are only 2 ME outcomes, #1 (p = prob. of success) #2 (1 - p = prob of failure) 4. probability of success is the same for each trial. p=0. # of success in n trials x=0,1,2,...,n.

Mean

Mx= n(trials) x p(probability)

At least 3 are wireless only

P(.3)= 1-p(<2)=1-.003=.997

Computing the probability of random variable X

P(zero success) = P(FFFF) = P(F) x P(F) x P(F) x P(F) P(one success) = P(SFFF or FSFF or FFSF or FFFS) = P(SFFF) + p(FSFF) + P(FFSF) + P(FFFS) Calculator 2nd Vars -> A:Binompdf (n (size), p, x (trial #)) Given answer exponent: 2.401^E -5= .00002401

Binomial Distributions

is a discrete probability distribution that describes probability for experiments with 2 mutually exclusive outcomes.

Empirical Rule (68-95-99.7)

np(1-p) greater than equal to 10 Anything outside of M-2sigma/ 2 st. dev is unusual <.05 is unusual 1. Draw a curve 2. plot Mx 3. Plot the # in the question whether higher or lower than Mx 4.higher than Mx -> M + 2sigma 5. lower than Mx -> M - 2sigma 6. If answer closer to Mx the # from question is outside of 2 st. dev : unusual 7. If answer is farther than Mx, # is not unusual within 2 st dev

standard deviation

sigmax = square root of np(1-p)