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## Study guide for exam 2

1 review
by: Emily Binakonsky

141

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5

# Study guide for exam 2 ENGR 0020: Probability and statistics for Engineers I

Emily Binakonsky
Pitt
Probability and Statistics for Engineers 1

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The material covered on chapters 5, 6, 8 and 9 is the conceptual. Equations are also given.
COURSE
Probability and Statistics for Engineers 1
PROF.
TYPE
Study Guide
PAGES
5
WORDS
KARMA
50 ?

1 review
Yilun Xu

## Popular in Engineering and Tech

This 5 page Study Guide was uploaded by Emily Binakonsky on Monday March 16, 2015. The Study Guide belongs to ENGR 0020: Probability and statistics for Engineers I at University of Pittsburgh taught by Maryam Mofrad in Spring2015. Since its upload, it has received 141 views. For similar materials see Probability and Statistics for Engineers 1 in Engineering and Tech at University of Pittsburgh.

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Date Created: 03/16/15
Informational review on chapters 5 6 8 amp 9 Emily Binakonsky I Chapter 5 1 Binomial Distributions To use the Bernoulli process the experiment must meet the following criteria a The experiment must have repeated trials b The outcome of each trial has to be either a success or failure c The probability of success p remains constant from trial to trial d The repeated trials must be independent of each other Binomial random variable is the number X of successes in n Bernoulli trials The binomial distribution is the probability distribution of the discrete random variable It s values will be denoted by bx n p bxnp 219an x x 012 The mean and variance are a np and 02 npq 2 Hypergeometric Distribution Must mean the following criteria a A random sample of size n is selected wo replacement from N items b Of the N items k is classified as successes and N k are classified as failures Hypergeometric random variable is the number X of successes of a hypergeometric experiment The Hypergeometric distribution is the probability distribution of the hypergeometric variable k N k hxNnk n max0 n N k S X S minn k N n nk The mean and variance are u W and 0392 N1 3 Negative Binomial Distribution bx k p Zihaqu x x kk 1k 2 4 Geometric Distribution gxp pa x 01 The mean and variance are u and 0392 12p 5 Poison Distribution Informational review on chapters 5 6 8 amp 9 Emily Binakonsky Poisson random variable is the number X of outcomes occurring during a Poisson experiment The Poisson Distribution is its probability distribution px3t T x 012 PO 315 320 2906 3t The Poisson Distribution mean and the variance are the same At Chapter 6 1 Continuous Uniform Distribution rectangular distribution The denistyh function of the continuous uniform random variable X on the interval AB is 1 A lt lt B fxAB B A x O elsewhere The mean and variance of the uniform distribution are A B d 2 B A2 2 aquot a 12 2 Normal Distribution Gaussian distribution it is the most important continuour probability distribution in the entire field of statistics The bellshaped curve is referred to the normal curve The normal random variable is a continuous random variable X having a bellshaped distribution The density of the normal random variable X with mean u and variance 02 is ixM2 e 202 00 lt x lt 00 1 quotOquot a ma Examining the 1st and 2quot I derivatives of nx u a the following properties are observed 1 The mode which is the the point on the horizontal axis where the curve is a maximum occurs at x u 2 The curve is symmetric about a vertical axis through the mean u 3 The curve has its points of in ection at x u i a it is concave downward if u a lt X lt u a and is concave upward otherwise A Standard Normal Distribution is the distribution of a normal random variable with mean 0 and variance 1 X M Z 039 Informational review on chapters 5 6 8 amp 9 Emily Binakonsky 3 IfX is a binomial random variable with mean u np and variance 02 npq X np then Z vnpq 4 Exponential Distribution x 1 fx 6 339 x gt 0 O elsewhere mean and variance respectively are u B and variance 02 82 Describe the relationship between the Poisson process and the exponential distribution and the gamma distribution Chapter 8 and 9 5 Chisquared Distribution Has one parameter v degrees of freedom The Chisquared distribution is an important special case of the gamma distribution It is obtained by letting OL v2 and B 2 where v is a positive integer The density function is given by the formula 21 5 fx 122 e 239 x gt 0 where v is a positive interger O elsewhere the mean u V and the varaince 02 212 6 TDistribution as n gt oo is the standard normal dist nz 01 Let X1X2XI1 be the independent random variables that are all normal with mean u and standard deviation 6 Let 1 n 2 1 n 2 Then the random variable T X has a tdistribution with Ti 1 n 1 desgrees of freedom Let tv denote the density function curve for V degrees of freedom 0 Each tv curve is centered at 0 and bellshaped 0 Each tv curve is spread out more than the standard normal curve 12 VT v 2 for v gt 2 where Ttv As z increases the tv curve spread decreases 0 As 1 00 the tv sequence approaches the standard normal curve 7 F Distribution The F statistic is defined to be the ratio of two independent chisquared random variables each divided by its degrees of freedom f F Informational review on chapters 5 6 8 amp 9 Emily Binakonsky where U and V are independent random variables having chisquared distributions v1 and 172 degrees of freedom How to write a foo v1 v2 for foo with 121 and 122 degrees of freedom 1 131072 121 if Si and 8 are the variances of independent random samples of size n1 and n2 taken from normal populations with variances oi and 0 respectively then f1 al71 172 withv1n1 1 andv2n2 1 8 Normal QuantileQuantile plot A plot of the ordered observations yi against the corresponding normal distribution s quantile 9 Sampling Statistics Distribution a Sampling Distribution of the difference between two means if independent samples of size n1 and n2 are drawn at random from two populations discrete or continuous with means 11 and a2 and variances and 012 and 022 respectively then the sampling distribution of the differences of means X1 X2 is approximately normally distributed with a Mean xi 22 M1 M2 2 2 Z 1 2 And variance 0X1X2 n1 quot2 Y Y Thus Z 1 2 M1 2 1s approx a standard normal var1able 0392 0392 are 9 Classic Estimation Methods take a sample find the sample mean and sample variance a Notation 1 6 denotes interest parameter b Unbiased Estimators A statistic 3 as am unbiased estimator of 6 if aeEi9 for all values of i9 The difference of Ei9 is called the bias of 3 Informational review on chapters 5 6 8 amp 9 Emily Binakonsky c Interval Estimation An interval estimate of the population parameter 6 is given by the formula 6 lt 0lt 61 Different pulled samples from the same population will give different values of the estimator 3 we can calculate a probability that a we will pick a sample that produces an interval containing the true population parameter 6 P Llt 6lt U1 a L U is is called the 1001 a Confidence Interval The confidence coefficient degree of confidence is 1 a The endpoints 613 and 61 are the confidence limits d Mean Estimation determining the confidence interval Step 1 Setup the probability statement to represent 1 a of Z values Step 2 ZZX Mmn gtRearrange for u Case 1 Use the Normal Distribution Case 2 Use the TDistribution Case 3 Use the ChiSquared Distribution Make sure to mention the Central Limit Theorem NOTE Redoing the previously assigned homework would be the best way to prep for the numerical aspect of the second exam I will upload all my homework solutions tomorrow and email them to you

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