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1 Prof. Indrajit Mukherjee, School of Management, IIT Bombay Example Probability distributions can be estimated from relative frequencies. # of televisions.

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Presentation on theme: "1 Prof. Indrajit Mukherjee, School of Management, IIT Bombay Example Probability distributions can be estimated from relative frequencies. # of televisions."— Presentation transcript:

1 1 Prof. Indrajit Mukherjee, School of Management, IIT Bombay Example Probability distributions can be estimated from relative frequencies. # of televisions # of household s XP(X) 01,21800.012 132,37910.319 237,96120.374 319,38730.191 47,71440.076 52,84250.0.28 101,5011.000 e.g. P(X=4)=P(4)=0.076 1,218/101,501=0.012

2 2 Prof. Indrajit Mukherjee, School of Management, IIT Bombay # of televisions # of household s XP(X) 01,21800.012 132,37910.319 237,96120.374 319,38730.191 47,71440.076 52,84250.0.28 101,5011.000 Example E.g. what is the probability there is at least one television but no more than three in any given household? “at least one television but no more than three” P(1 ≤ X ≤ 3) = P(1) + P(2) + P(3) =.319 +.374 +.191 =.884

3 3 Prof. Indrajit Mukherjee, School of Management, IIT Bombay Example Find the mean, variance, and standard deviation for the population of the number of color televisions per household… (from Example 7.1) # of televisions# of households XP(X) 01,21800.012 132,37910.319 237,96120.374 319,38730.191 47,71440.076 52,84250.0.28 101,5011.000

4 4 Prof. Indrajit Mukherjee, School of Management, IIT Bombay # of televisions # of households XP(X) 01,21800.012 132,37910.319 237,96120.374 319,38730.191 47,71440.076 52,84250.0.28 101,5011.000 Example Find the mean, variance, and standard deviation for the population of the number of color televisions per household…

5 5 Prof. Indrajit Mukherjee, School of Management, IIT Bombay 11760.05-1.64 21830.15-1.04 31850.25-0.67 41900.35-0.39 51910.45-0.13 61920.550.13 72010.650.39 82050.750.67 92140.851.04 102200.951.64 Ten observations on the effective service life in minutes of batteries used In a portable personal computers are as follows: 176,191,214,220,205, 192,201,190,201,190,183,185. we hypothesize that battery life is adequately model by normal distribution. To use probability plotting to investigate this hypothesis. First arrange the observations in ascending Order and calculate their cumulative frequencies (j-0.5)/10 as shown in a table Probability Plots Example

6 6 Prof. Indrajit Mukherjee, School of Management, IIT Bombay 1860.05-1.64 2870.15-1.04 387.20.25-0.67 487.40.35-0.39 587.80.45-0.13 688.20.550.13 788.90.650.39 889.60.750.67 989.70.851.04 10900.951.64

7 7 Prof. Indrajit Mukherjee, School of Management, IIT Bombay 1.230.49 1.080.590.28 0.180.10.94 1.330.191.160.980.34 0.190.210.4 0.040.830.050.630.340.750.040.530.43 0.0440.810.150.560.840.870.490.520.25 1.20.710.190.410.50.561.10.650.27 0.50.770.730.340.170.160.27 An article in the 1993 volume of the transactions of the American fisheries society reports the result of a study to investigate the merry contamination In large mouth bass. A sample of fish was selected from 53 Florida lakes And mercury concentration in the muscle tissue was measured (ppm). The mercury concentration values are Normal Distribution Example

8 8 Prof. Indrajit Mukherjee, School of Management, IIT Bombay k0.010.050.10.20.250.30.40.50.60.70.750.80.90.950.99 00.90440.59870.34870.10740.05630.02820.0060.0010.0001000000 10.99570.91390.73610.37580.2440.14930.04640.01070.00170.000100000 20.99990.98850.92980.67780.52560.38280.16730.05470.01230.00160.00040.000100 310.9990.98720.87910.77590.64960.38230.17190.05480.01060.00350.0009000 410.99990.99840.96720.92190.84970.63310.3770.16620.04830.01970.00640.000100 5110.99990.99360.98030.95270.83380.6230.36690.15030.07810.03280.00160.00010 61110.99910.99650.98940.94520.82810.61770.35040.22410.12090.01280.0010 71110.99990.99960.99840.98770.94530.83270.61720.47440.32220.07020.01150.0001 8111110.99990.99830.98930.92360.85070.7560.62420.26390.08610.0043 91111110.99990.9990.9940.97180.94370.89260.65130.40130.0956 Binomial Table… The probabilities listed in the tables are cumulative, i.e. P(X ≤ k) – k is the row index; the columns of the table are organized by P(success) = p

9 9 Prof. Indrajit Mukherjee, School of Management, IIT Bombay k0.010.050.10.20.250.30.40.50.60.70.750.80.90.950.99 00.90440.59870.34870.10740.05630.02820.0060.0010.0001000000 10.99570.91390.73610.37580.2440.14930.04640.01070.00170.000100000 20.99990.98850.92980.67780.52560.38280.16730.05470.01230.00160.00040.000100 310.9990.98720.87910.77590.64960.38230.17190.05480.01060.00350.0009000 410.99990.99840.96720.92190.84970.63310.3770.16620.04830.01970.00640.000100 5110.99990.99360.98030.95270.83380.6230.36690.15030.07810.03280.00160.00010 61110.99910.99650.98940.94520.82810.61770.35040.22410.12090.01280.0010 71110.99990.99960.99840.98770.94530.83270.61720.47440.32220.07020.01150.0001 8111110.99990.99830.98930.92360.85070.7560.62420.26390.08610.0043 91111110.99990.9990.9940.97180.94370.89260.65130.40130.0956 Binomial Table… “What is the probability that Pat gets no answers correct?” i.e. what is P(X = 0), given P(success) =.20 and n=10 ? P(X = 0) = P(X ≤ 0) =.1074

10 10 Prof. Indrajit Mukherjee, School of Management, IIT Bombay k0.010.050.10.20.250.30.40.50.60.70.750.80.90.950.99 00.90440.59870.34870.10740.05630.02820.0060.0010.0001000000 10.99570.91390.73610.37580.2440.14930.04640.01070.00170.000100000 20.99990.98850.92980.67780.52560.38280.16730.05470.01230.00160.00040.000100 310.9990.98720.87910.77590.64960.38230.17190.05480.01060.00350.0009000 410.99990.99840.96720.92190.84970.63310.3770.16620.04830.01970.00640.000100 5110.99990.99360.98030.95270.83380.6230.36690.15030.07810.03280.00160.00010 61110.99910.99650.98940.94520.82810.61770.35040.22410.12090.01280.0010 71110.99990.99960.99840.98770.94530.83270.61720.47440.32220.07020.01150.0001 8111110.99990.99830.98930.92360.85070.7560.62420.26390.08610.0043 91111110.99990.9990.9940.97180.94370.89260.65130.40130.0956 Binomial Table… “What is the probability that Pat gets no answers correct?” i.e. what is P(X = 0), given P(success) =.20 and n=10 ? P(X = 2) = P(X≤2) – P(X≤1) =.6778 –.3758 =.3020 remember, the table shows cumulative probabilities…

11 11 Prof. Indrajit Mukherjee, School of Management, IIT Bombay Sampling Distribution of Proper analysis and interpretation of a sample statistic requires knowledge of its distribution. Calculate to estimate µ Population µ (Parameter) Select a random sample Sample (Statistic) Process of Inferential statistics

12 12 Prof. Indrajit Mukherjee, School of Management, IIT Bombay Population (mean, μ, is unknown) Estimation Process Random Sample I am 95% confident that μ is between 40 & 60. Sample Mean = 50

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