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0.01 for 10 ?x?20. The mean of the distribution is 15 and the standard deviation is 2.89. Find the
probability that x falls between 12 and 15. Can you please show your steps.
Solution
First we convert the random variable into a standard normal variable so that we can
get the results from the z-table. z=(x-mean)/standard_deviation Thus we need to find the
probability that our z-variable lies between (12-15)/2.89 and (15-15)/2.89, or -1.038(-1.04) and
0. From cross-checking the z-table http://lilt.ilstu.edu/dasacke/eco148/ztable.htm We find that
probability of z-variable lying below -1.04 is 0.4443 or 44.43% Thus the probability of it lying
between -1.04 and 0 is (0.5-0.4443) = 0.0557 or 5.57% (Note that there is a total 50%
proabability of z-variable lying below 0.) This can also be shown as P(-1.04-1.04)+P(z>0)=-
0.4443+0.5=0.0557

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0.01 for 10 x20. The mean of the distribution is 15 and the standa.pdf

  • 1. 0.01 for 10 ?x?20. The mean of the distribution is 15 and the standard deviation is 2.89. Find the probability that x falls between 12 and 15. Can you please show your steps. Solution First we convert the random variable into a standard normal variable so that we can get the results from the z-table. z=(x-mean)/standard_deviation Thus we need to find the probability that our z-variable lies between (12-15)/2.89 and (15-15)/2.89, or -1.038(-1.04) and 0. From cross-checking the z-table http://lilt.ilstu.edu/dasacke/eco148/ztable.htm We find that probability of z-variable lying below -1.04 is 0.4443 or 44.43% Thus the probability of it lying between -1.04 and 0 is (0.5-0.4443) = 0.0557 or 5.57% (Note that there is a total 50% proabability of z-variable lying below 0.) This can also be shown as P(-1.04-1.04)+P(z>0)=- 0.4443+0.5=0.0557