Question 1076610: A)If a normal distribution has a mean of 343.11 and standard deviation of 110.24, what is P(x≤213.98)?
B)If a normal distribution has a mean of 17 and standard deviation of 4 what is P(x≤22)?
C)For a standard normal distribution with mean of 0 and standard deviation of 1, what is P(x≤0.4)?
I am having problems with this assignment I have several others that are similar to this question. Can I please get help.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd. The concept of a test statistic being the (value-mean)/std dev appears everywhere in statistics.
want z < (213.98-343.11)/110.24=-1.171
want area less than that for z. Go to calculator (see below) want left end--some use -9999, but -6 is a far enough sd to take in the left end, then comma then the answer you have here. Hit enter, and the probability appears.
probability is 0.1207
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want z<1.25 which is 5/4 sd above the mean
0.8944 is probability
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Standard normal distribution
z=(x=mean)/sd=(0.4-0)/1=0.4
want the z less than 0.4 This can be found in a table or on a calculator
2nd + VARS+2+(-6,0.4)+ENTER, and probability is 0.6554.
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