Normal distribution table values

  • Thread starter aurao2003
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Homework Statement



Hi
I need a clarification before my S2 exam on Thursday. How do I find a value of P(Z<0.7925)? This is bearing mind that all the table values are to 2 decimal places.
Thanks.

Homework Equations


Normal Distribution Tables



The Attempt at a Solution


Previously, I have compare it to the closest value and then subtract to get my final result. For example, if I need P(Z<1.212), I would compare it to P(Z<1.21) and P(Z<1.22). Since it is closer to 1.21, I would subtract .002 from the table value. Is this method correct? I am getting a slightly different result to the questions. Please help.
 

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  • #3
HallsofIvy
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Homework Statement



Hi
I need a clarification before my S2 exam on Thursday. How do I find a value of P(Z<0.7925)? This is bearing mind that all the table values are to 2 decimal places.
Thanks.

Homework Equations


Normal Distribution Tables



The Attempt at a Solution


Previously, I have compare it to the closest value and then subtract to get my final result. For example, if I need P(Z<1.212), I would compare it to P(Z<1.21) and P(Z<1.22). Since it is closer to 1.21, I would subtract .002 from the table value. Is this method correct? I am getting a slightly different result to the questions. Please help.
Surely that's not what you meant to say! Using this table: http://www.math.unb.ca/~knight/utility/NormTble.htm [Broken]
I get that P(z< 1.21)= 0.8869 and P(z< 1.21)= 0.8888. You certainly don't want to subtract anything from P(z< 1.21) because the probability is increasing, not decreasing. And if you were to add .002, you would get 0.8908 which is already past 0.8888.

Instead, note that 0.8889- 0.8868= .0019. Since 1.212 is .2= 1/5 of the way from 1.21 to 1.22, add 1/5 of that difference: .2(.0019)= 0.00038. P(z< 1.212) will be approximately 0.8889+ .00038= 0.88728.

Similarly to approximate P(z< 0.7925), I note in the table that P(z< 0.79)= 0.7852 and P(z< 0.80)= 0.7881. That is a difference of 0.7881- 0.7852= 0.0029. 0.7925 is .25= 1/4 of the way from 0.79 to 0.80 so I find 1/4 of that difference- .0029/4= .0029(.25)= 0.000725. The linear interpolation from this table for P(z< 0.7925) is 0.7852+ 0.000725= 0.785925.

That is, as SteamKing said, a "linear interpolation"- we are assuming the graph is linear between 0.79 and 0.80 which is only approximately correct.
 
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