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Unformatted text preview: Business Statistics
Fifth Edition Ken Black Chapter 6: Continuous Distributions These notes are not to be reproduced without the written permission of RB. Alt 6] 6.0 Continuous Random Variables in General 0 A continuous random variable can take on values at every point over a given
interval. (Page 142) @ Real no. line 10:45 1 1:00 (Time at which you arrive for class) 0 Examples of continuous random variables:
0 Stock market returns
0 Quality characteristics of ﬁnished products (such as net contents)
0 Heights of males; heights of females
0 Age at time of death
0 Features of a continuous random variable (Page 183)
o The possible values are uncountable.
o The probability that the random variable takes on a speciﬁc value is 0.
0 Only an interval of values has a nonzero probability. 0 The entire area under the curve is equal to l. o The probability for an interval of values will be shown as the area under the distribution. f(x) These notes are not to be reproduced without the written permission of RB. Alt 6—2 0 Details 0 It doesn’t matter whether endpoints are included in the interval b/c: Pb<X<M=Pb£X<H=Pb<X5H=Pb£XSH
Why? P[X = a] = P[X = b] = #. 0 Data are IICVGI' COHtlI’lIlOUS! 0 Comparison of Probability Distributions for Discrete and Continuous R.V.’s Mm P(X) P(a Sbe) LX These notes are not to be reproduced without the written permission of EB. Alt 63 6.2 The Normal Distribution
6.2.1 The Standardized Normal Distribution 0 The probability distribution of a standard normal random variable Z is shown below: @ g A—@ ' 13(2): #z=0
V(Z)=GZZ=1 0 Other Properties:
0 Total area under the curve is l. o The curve is symmetric around 0
o P(Z > 0) = 0.5
¢ Finding probabilities for a standard normal random variable: 0 Use Table A.5 (gives area from 0 to a righthand value 2) — Page 788
0 Use Minitab II In the worksheet, input the value of Z in any column, e.g., Cl
I Calc > Probability Distributions > Normal I Select Cumulative Probability to obtain P(Z < 2) These notes are not to be reproduced without the written permission of RB. Alt 64 Exercise:
Suppose that Z is a standard normal random variable. 0 Find P(ZS 2.42). fgz) P(Z s 2.42) l : = 0.5 — .4922 (from Table A.5)
l r “ __ = 0.0078 igajﬂ'lflllill , ,,,,,,,7,,Wiim, massaim" ,,,, ,7, a ,WVQTT? 242 0
I From Minitab:
Normal with mean = 0 and standard deviation = 1
x P( X <= x )
—2.42 0.0077603
0 Find P(—1.07_<_ Z_<_ 2.33)
Z
ff ) E] P(1.07S z: 2.33) =P(1.07S ZS 0) +P(OS 23 2.33) i i ( from Table A.5)
l.07 0 2.33 = 0.8478 These notes are not to be reproduced without the written permission of RB. Alt 6—5 Exercise:
For a gard normal random variable Z, ﬁnd the value k such that: o P(ZZ k) = .01 z
ff ) From Table A.5, P(0 : Z s 2.3% 0.4901 P(ZZ 2.33) = :>k= These notes are not to be reproduced without the written permission of PB. Alt 6—6 6.2.1 The Normal Distribution in General . E]
f (X) 0 The probability distribution is moundshaped.
o u is the expected value of the distribution.
0 o is the standard deviation of the distribution. 0 The equation of the cg (page 189) is not of interest to us. 0 Standardize X to ﬁnd areas under the normal curve of X. Xv!
0' Z: {Procedure for standardizing X} 0 We can now use Table A5 to calculate probabilities for X o The standardized variable Z measures how many standard deviations X is above or below its mean. These notes are not to be reproduced without the written permission of EB. Alt 67 Examgle: A potato chip packaging plant has a processing line that ﬁlls 12 ounce bags of
potato chips. At the current setting of the machine, the quality control engineer
knows that the actual distribution of weights in the bags follows a normal
distribution with a mean of 12.0 ounces and a standard deviation of 0. I 8 ounces. a) What percentage of all bags ﬁlled contain exactly 12 ounces? P(X: 131E] b) What percentage of all bags ﬁlled contain more than 12.4 ounces? P(X >12.4)
fm 12 4 12
2 Hz > 4;)
A 0.18
= P(Z > 2.22)
@0132
J a X These notes are not to be reproduced without the written permission of PB. Alt 6—8 c) Management is concerned when 12—ounce bags of potato chips contain less than
11.75 ounces. The quality control engineer can set the ﬁlling machine so that
actual mean ﬁlling weight is whatever he chooses, but the standard deviation
always remains at 0.18 ounces. What mean ﬁlling weight should he set the machine to if he wants only 1% of all bags to contain less than 11.75 ounces? Find ,u so thatP (X< 11.75): .01 Normal Probability Distribution
Mean: ?, StDev=0.18 11.75—[1 0.18 ) .01: P(X <ll.75)= P(Z < .01: P(Z < —2.33)from Table 3 saw: 233 0.18 ,u = 12.17 ounces These notes are not to be reproduced without the written permission of RB. Alt 69 ...
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 Fall '08
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 Normal Distribution, Standard Deviation, probability density function, potato chips, written permission

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