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anonymous

  • one year ago

Find the area under the normal distribution curve between z = 1.30 and z = 3.25.

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  1. jim_thompson5910
    • one year ago
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    which option do you want to use: table or calculator?

  2. anonymous
    • one year ago
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    table

  3. jim_thompson5910
    • one year ago
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    ok we can use a table like this one https://www.stat.tamu.edu/~lzhou/stat302/standardnormaltable.pdf

  4. jim_thompson5910
    • one year ago
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    does something like that look familiar?

  5. anonymous
    • one year ago
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    Yes Thanks I'm kind of confused with this lesson

  6. jim_thompson5910
    • one year ago
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    first we'll find P(Z < 1.30) look at the row that starts with `1.3` then find the column that has `.00` at the top of the column what number is at the intersection of that row and column?

  7. anonymous
    • one year ago
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    .90320

  8. jim_thompson5910
    • one year ago
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    good

  9. jim_thompson5910
    • one year ago
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    do the same for P(Z < 3.25)

  10. jim_thompson5910
    • one year ago
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    look at the row starting with `3.2` and the column starting with `0.05`

  11. anonymous
    • one year ago
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    .99942

  12. jim_thompson5910
    • one year ago
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    now subtract the two (bigger - smaller)

  13. anonymous
    • one year ago
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    wow Thanks a lot. That's an answer choice I appreciate it

  14. jim_thompson5910
    • one year ago
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    you should get .99942-.90320 = 0.09622 so P(1.30 < z < 3.25) = 0.09622 The rule I used was \[\Large P({\color{red}{a}} < z < {\color{blue}{b}}) = P(z < {\color{blue}{b}}) - P(z < {\color{red}{a}})\] where b > a so, \[\Large P({\color{red}{1.30}} < z < {\color{blue}{3.25}}) = P(z < {\color{blue}{3.25}}) - P(z < {\color{red}{1.30}})\]

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