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The heights of 18 year old men are apprx normally distrubuted,with mean 68 inche

ID: 2913529 • Letter: T

Question

The heights of 18 year old men are apprx normally distrubuted,with mean 68 inches and standard deviaton 3 inches (based oninformation from Statistical Abstract of the United States, 112thEdition). a. What is the probability that an 18year old man selected atrandom is between 67 and 69 inches tall? b. If a random sample of nine 18 year ol men is selected, whatis the probability that the mean height _ is between 67 and 69inches?                                                                                                             x c. Compare our answers for parts a and b. Is the probabilityin part b much higher? Why would you expect this? The heights of 18 year old men are apprx normally distrubuted,with mean 68 inches and standard deviaton 3 inches (based oninformation from Statistical Abstract of the United States, 112thEdition). a. What is the probability that an 18year old man selected atrandom is between 67 and 69 inches tall? b. If a random sample of nine 18 year ol men is selected, whatis the probability that the mean height _ is between 67 and 69inches?                                                                                                             x c. Compare our answers for parts a and b. Is the probabilityin part b much higher? Why would you expect this?

Explanation / Answer

The heightsof 18 year old men are approx normally distributed, with mean 68inches and standard deviation 3 inches (based on information fromStatistical Abstract of the United States, 112thEdition).
     {Mean} = = 68
     {Std Dev} = = 3
     {NormalZ Distribution} ---->   Z = (x- )/
     {Normal Z Distribution Of SampleMean} ---->   Z = (x -)/(/sqrt(n))


a. What isthe probability that an 18 year old man selected at random isbetween 67 and 69 inches tall?
    = Prob{((67) - (68))/(3) < Z < ((69) -(68))/(3)}
     = Prob{(-0.3333) < Z < (0.3333)}
     = 0.2611


b. If arandom sample of nine 18 year old men is selected, what is theprobability that the mean height _ is between 67 and 69inches?             
    = Prob{((67) - (68))/(3/sqrt(9)) < Z < ((69) -(68))/(3/sqrt(9))}
     = Prob{(-1.0) < Z < (1.0)}

    = 0.6827
                                                                                              
c. Compareanswers for a and b. Is the probability in part b muchhigher? Why would you expect this?
     {It is greater for "b" because Std Dev "" is smaller andthere is LESS spread
        about the Mean, therebyenclosing MORE area between values (67) &(69).}


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