normal distribution question

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davidh

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The number of automobile accidents per week in Pascalville is normally distributed with a mean of 16.3 accidents and standard deviation of 2.7 accidents. What is the probability that there will be at least 12 automobile accidents in a given week?

ok so! what i have is standard deviation is 2.7 and mean is 16.3 so my thought process is 12-16.3/2.7 = -4.3/2.7 = -1.592
im pretty sure im supposed to get a number less then 0 correct and if so what did i do wrong?
 
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1st of all 12 - 16.3/2.7 is not the same as (12-16.3)/2.7
Then you lost me. You said that you are supposed to get a number less than 0 (and you did since -1.592 < 0) and if so what did I do wrong. You got what you thought you should get ( a negative number) but think that is wrong. Very strange!

All z-values to the left of the mean are negative and 12 is to the left of 16.3.

Now you need to convert z= -1.592 to an area and depending on your z-table manipulate this area to correspond to the area you want.
 
Jomo said:
1st of all 12 - 16.3/2.7 is not the same as (12-16.3)/2.7
Then you lost me. You said that you are supposed to get a number less than 0 (and you did since -1.592 < 0) and if so what did I do wrong. You got what you thought you should get ( a negative number) but think that is wrong. Very strange!

All z-values to the left of the mean are negative and 12 is to the left of 16.3.

Now you need to convert z= -1.592 to an area and depending on your z-table manipulate this area to correspond to the area you want.
Click to expand...
yes thats my bad
i now am stuck on 1-p(z=-1.592)
i got here by doing =px greater than or equal to 12)
=p(z ≥ -1.592)
=1-P(Z=0-1.592)
unsure what to do now
 
actually i forgot to use the 9 so it would be 94.4% rounded and i forgot that its 0.0 not 0.
 
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