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Week 6 Bstats notes

by: Sophie Levy

Week 6 Bstats notes MATH 1140

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Sophie Levy

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All week 6
Business Statistics
Robert Herbert,
Class Notes
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This 10 page Class Notes was uploaded by Sophie Levy on Sunday October 9, 2016. The Class Notes belongs to MATH 1140 at Tulane University taught by Robert Herbert, in Summer 2015. Since its upload, it has received 2 views. For similar materials see Business Statistics in Math at Tulane University.


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Date Created: 10/09/16
Week 6 10/3 Section 4.3 continued Binomial Experiment review 1. N trials 2. each trial has 2 outcomes (success and failure) 3. p=success and q=p-q=failure 4. trails are independent - x=number of successes - x=0,1, 2,…n where n is the number of trials n x n-x - p(x) = ( )xp )(q ) because we can’t use calculator we will use table on exam Theorem: if X is binomial, the expected value of: - E(X) = Σx*p(x)= np - V(X)= Σ(x-u) p(x) = npq - Indicator function is only for binomial random variables o X=x +x 1+x 2 n o x= jth indicator function o E(x)=1jp + 0*q = p o V(x)= 1 *p + 0 *q – p 2 j  =p-p= p(1-p)  =pq - in a binomial experiment, the indicator functions are independent, therefore the variance of the sum is the sum of the variances o V(X)= V(x ) +1V(x )…+V2x ) n  =pq + pq…+pq o Standard deviation = square root of npq Example: suppose 40% of New Orleans residents are registered voters; if we obtain a sample of 50 residents, what is the expected number of registered voters? - 2 paths: registered or not registers - success: p: registered - failure: q: not registered - n=50 - E(X)= 50(.4)= 20 - V(X)= 50(.4)(.60)=12 Section 4.4 Hypergeometric ONLY: sampling without replacement (similar to a binomial experiment except it fails the 4 condition– the events of success will not be independent) Ex: Urn problem with 8 red and 12 yellow; we select 7 balls without replacement; fine the probability that we get 4 red and 3 yellow? - samples= ( 207)= 77,520 o the number of samples with 4 red and 3 yellow is given as 8 12 ( 4*( 3)=15,400 - p(4 red and 3 yellow) = 15,400/77,520= .1986 Ex: - N=20total number in population - n=7 sample size - k=8 number of successes popular - N-k=12 number of failures possible - x=2 actual number of successes - n-x=5 actual number of failures - ( x*(N-kn-x / (n) - E(X)= n(k/n) - V(X)=(N-n/N-1) * n(k/n) *(1-k/n) o Finite connection factor: N-n/N-1 Handout Ex #2: From a group of 15 engineers, 10 are randomly selected for employment by a company. - N=15 - n=10 a) What is the probability that the 10 will include the 5 most qualified engin15rs? o ( 10)= 3003 o ( )5* ( 15) = 1*252 o p(5mQ)= 252/3003= .0839 b) What is the probability that none of the 10 selected individuals are most qualified? 5 10 15 o ( )0*( 10) / ( 10) o 1 * 1 / 3003 o 1/3003! Very unlikely c) What is the probability that at least two of the most qualified engineers are hired? o “At least two” means up to 5 since there is only 5... difficult o Instead, find the complement: o 1 - ( )0 110)/3003 – ( )1 109/3003 There are instances where its hypergeometric but we treat it as binomial: the 40% New Orleans residents example– we treat it as a binomial because the (N) sample size is large and n is of a modest size. There is no practical difference between sampling with replacement and sampling without replacement. It is SO unlikely to pick a single element in the first place!! - Use binomial when n/N < .05 Week 6 10/5 Section 4.6: the normal distribution The normal distribution with value ��� and standard deviation σ, is the curve defined by the equation: This is defined for the parameters μ, which can be any real number, and ���, which can be any positive real number. If we set ��� = 0 and ��� = 1, then we get the standard normal curve, which equals .4: - It will be a bell curve– this is the important part - So normal cure = bell curve - We refer to it as a normal curve because it appears in such great frequency - Function properties: o The domain all real numbers o Are nonnegative for all x o The area between the graph of the function and the x-axis is 1. If we say that a random variable, X, is normally distributed with a mean value, ���, and standard deviation, σ, this means that we may compute probabilities for X as areas beneath the curve - Given values “a” and “b”, we can ask what’s the probability that the random variable falls between a and b: prob(a ≤ x ≤ b) - “Slogan is: probability = area” If “z” has a standard normal distribution, then we can compute probabilities for z as areas beneath , - area = prob(a ≤ z ≤ b) - the probability wil be whatever area is beneath the curve and above the axis - z often denotes a random variable with the standard normal distribution Example 15 (from handout): Suppose the random variable Z has the standard normal distribution. Find the following: a) The probability that Z is at most 1.23 - Prob(z ≤ 1.23) - If we believe that z has a standard normal distribution, we can look at the table: o Right body area: area to the right of the middle of the curve, 0 o We look at the table to find what it is at 1.23 o We round of the z to 2 decimal places  We get up to the tenths place on the y axis  We get the hundredths place on the x axis  Table tells us 0.3907 - We add .5 to .3907 to get 0.8907 b) The probability that Z is at least 2.13 - Prob(z ≥ 2.13) o We want to calculate the area greater than 2.13 - The table tells us the right body area in between 0 and 2.13 o On the table, 2.13 is 0.4834 - We subtract because we are finding the difference: o 0.5-0.4834= 0.0166 c) The probability that Z takes a value between 1 and 2.5 - Prob(1 ≤ z ≤ 2.5) - We can calculate the RBA 0 to 2.5 o Table tells us its 0.4938 - We can calculate the RBA from 0 to 1 o Table tells us 0.3413 - Prob(z)=.4938-.3413= .1523 d) The probability that Z takes a value between -1 and 5.2 - Prob(-1 ≤ z ≤ 5.2) - We find 0 to 1 o Table is 0.3413 - We want 0 to 5.2… not on the table– but notice that 3.99 is .49997, which is essentially .5– we can assume it is .5! o So table is .5 - Prob(z) = .3413 + .5 = .8413 Inverse problems: we use the table backwards th e) Find a 75 percentile for Z, that is , find a value c so that there is a 75% chance that Z is less than c and a 25% chance that Z is greater than c. - Area to left of c: 75% - Areas to the right of c: 25% - We assume that the RBA between 0 and c is .25 (because it has to add up to .5) o So then, we can find 0.25 IN THE TABLE– closest we can get is 0.2486 (.67) and 0.2517 (.68) o He wants us to use the one that is closest: 0.67 o C= approximately .67 f) Find a 95% confidence interval for Z, that is, find and interval centered at 0 so that there is a 95% chance that Z will take a value within this interval. - Paraphrase: find b so that prob(-b ≤ z ≤ b) =.95 - We cannot use table yet because the .95 is not a RBA o We use the positive number: we know RBA between 0 and b is half of .95: .4750 - BACKWARDS TABLE we see .4750 is 1.96, exactly! “the most famous number in statistics” - So b=1.96 - So we are 95% confident that z will fall within the range -1.96 ­ Inverse problem!!! ­ “That is, find a value ‘c’ so that p(x ≥ c ) =.1 so the p(x ≤ c) = .9 o if c is at .9, we want to find the area to the right of that o we know that there is .1 after c, so the RBA before c is .4 ­ find z score of c o in the table, the closest we get to a RBA of .4000 is .3997  the z score that corresponds to that is 1.28 ­ finding c o the z score formula tells us that (c-45,000)/5000 = 1.28 o c = 51,400 ­ the upper 10%: mileage of 51,400 or more d) Find the third quartile Q th 3 ­ Q 3eans 75 percentile o So p(x ≤ Q ) =3.75 and p(x ≥ Q ) = .25 3 ­ We want to find the RBA at .25 ­ Z score formula of .25 = .67 e) Find an interval of mileages centered at 45,000 so that there is an 85% chance that a truck’s mileage falls within this range. Note: such an interval is referred to as an 85% confidence interval. ­ Find interval centered at 45,000 ­ Because probability equals area, we can specific that the area between interval from a to  be is .85 o This is NOT the RBA o Divide .85/2 to find RBA o RBA= .425 ­ Z score of RBA of .4250 is 1.44 ­ Finding b o (b­45,000)/5,000 = 1.44 o b= 45,000 + 72,000 = 52,200 o a= 45,000 – 7,2000 = 37,800 ­ so  37,800 ≤ x ≤ 52,200 f) Find the probability that a truck is driven between 30,000 and 55,000. σ


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