The distribution of the number of occurrences of the letter t on the pages of a book is found to be a normal distribution with

The distribution of the number of occurrences of the letter t on the pages of a book is found to be a normal distribution with a mean of 44 and a standard deviation of 18. If there are 500 pages in the book, which sentence most closely summarized the data

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  1. The answer to this question is gotta be 0

    Step-by-step explanation:

    Because a normal distribution gives an area between 0 and z and the area for P(X=46) = 0

  2. The distribution of the number of occurrences of the letter t on the pages of a book is found to be a normal distribution with a mean of 44 and a standard deviation of 18. If there are 500 pages in the book, which sentence most closely summarizes the data?

    A. The letter t occurs less than 26 times on approximately 170 of these pages. 

    B. The letter t occurs less than 26 times on approximately 15 of these pages. 

    C. The letter t occurs more than 26 times on approximately 420 of these pages. 

    D. The letter t occurs more than 26 times on approximately 80 of these pages.


    mean = 44 
    sd = 18 
    that means that "26" is 1 s.d. down, or at the 16th %ile 

    so, there is a .16 chance that "t" will occur less than 26 times on any single page. 
    consequently, there is a .84 chance that it will occur more than 26 times on any single page. 

    Using that information, and knowing that 16% of 500 is 80, and 84% of 500 is 420, can you see where "C" is correct? 

  3. A normal distribution has a mean of 44 and a standard deviation of 8,

    We need to find z-score when x=50

    To find z-score we use formula

    [tex]z=\frac{x-mean}{standard deviation}[/tex]

    mean = 44

    standard deviation = 8 and x= 50

    [tex]z=\frac{50-44}{8}[/tex]

    [tex]z=\frac{6}{8}[/tex]

    [tex]z=\frac{3}{4}=0.75[/tex]

    the z-score for a value of 50 is 0.75

  4. 0.75

    Step-by-step explanation:

    Z-score tells the no. of standard deviations a data point is from the mean.

    the formula for z score is given by

    z-score= (data value - mean) / standard deviation

    Given:

    mean=44

    standard deviation= 8

    data point=50

    putting the values of mean(44), standard deviation(8) and given data value(50) in the above equation

    z-score= (50-44)/8

               = 6/8

               =0.75 !

  5. Here is the answer to the question.

    [tex]If a normal distribution has a mean of 44 and a standard deviation of 8,what is the z-score for a va[/tex]

  6. 244 - 165 = 79 . . . . one standard deviation
    323 - 244 = 79 . . . . one standard deviation

    The range of scores is ±1 standard deviation from the mean. The empirical rule says 
       68% of scores lie in that range.

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