Question 1159400: A petrol station sells, on the average, 14500 litters of petrol, per day with a standard deviation of 2500 litters. If a manager stocks 20,000 liters on a particular day, what is PROBABILITY that:
(i) At most 20,000 litters will be sold. (ii) More than 10,000 litters will be sold.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean is 14,500 liters of petrol sold per day.
standard deviation is 2,500 liters.
manager stocks 20,000 liters on a particular day.
quwstions are:
(i) At most 20,000 litters will be sold.
(ii) More than 10,000 litters will be sold.
a normal distribution is assumed.
let x = the number of liters sold for that particular day.
if you have the right calculator, you can solve this directly without having to find the z-score and look up the z-score probability tables, etc.
using my calculator, i enter the mean of 14,500 and the standard deviation of 2,500 and get:
p(x < 20,000) = .9860966011.
p(x > 10,000) = .9640697345.
using the online z-score calculator found at , i get:
p(x < 20,000) = .9861
p(x > 10,000) = .9641
the pictures of those results are shown below:
if i had to to it the hard way, i would use the z-score tables in the following manner.
z-score formula is z = (x - m) / s
z is the z-score
x is the raw score
m is the raw mean
s is the standard deviation
m = 14,500
s = 2,500
for p(x < 20,000), i would get z = (20,000 - 14,500) / 2,500 = 2.2
p(x < 20,000) becomes p(z < 2.2)
p(z < 2.2 means you are looking for area to the left of z = 2.2 in the z-score table that i used.
that table can be found at https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
looking in that table, you would find that the area to the left of a z-score of 2.2 = .98610.
that the probability that z < 2.2.
it's the same as i got for p(x < 20,000) with my calculator and the online calculator.
for p(x > 10,000), i would get z = (10,000 - 14,500) / 2,500 = -1.8
p(x > 10,000) becomes p(z > -1.8).
looking into the z-score table, i would find that the area to the left of a z-score of -1.8 = is .03593.
since i'm looking for the area to the right of that z-score, i would subtract that from 1 to get 1 - .03593 = .96407.
that's the probability that z > -1.8.
it's the same as i got for p(x > 10,000) with my calculator and the online calculator.
when you look for the z-score, the mean becomes 0 and the standard deviation becomes 1.
using z-scores rather than raw scores, the online calculator would show the following results.
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