SOLUTION: On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the variable is normally distributed. Find the probability that

Algebra ->  Probability-and-statistics -> SOLUTION: On the daily run of an express bus, the average number of passengers is 48. The standard deviation is 3. Assume the variable is normally distributed. Find the probability that       Log On


   

Question 747311: On the daily run of an express bus, the average number of passengers is 48. The
standard deviation is 3. Assume the variable is normally distributed. Find the probability that the bus will have fewer than 42?

more than 48?
Found 2 solutions by stanbon, mikeebsc:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
On the daily run of an express bus, the average number of passengers is 48. The
standard deviation is 3. Assume the variable is normally distributed.
Find the probability that the bus will have
a) fewer than 42?
z(42) = (42-48)/3 = -2
P(x < 42) = P(z < -2) = 0.0228
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b) more than 48?
z(48) = (48-48)/3 = 0
P(x > 48) = P(z > 0) = 0.5000
===================================
Cheers,
Stan H.
=================

Answer by mikeebsc(26) About Me  (Show Source):
You can put this solution on YOUR website!
For this we have to understand that average and "mean" are the same thing so:
LESS THAN 42:
We must first find he z-score using the formula: x-mean/Std.Dev. **or**
(42-48)/3=-2
Now we find the corresponding z score to -2.0 using the standard normal distribution table which = .0228 so:
P(x<42)=.0228
If you are allowed to use a calculator: using a TI-83 or TI-84 the steps are:
2nd DISTR
scroll down to normalcdf
lowerbound=-1,000,000 *this represents a number large enough to approximate - infinity as we are looking for EVERYTHING under 42 (make sure to use a neg sign and not the minus sign)
upperbound=42
mean=48
Std. Dev.=3
Press enter twice and you get .02275, or .0228; the same as if you had used the standard normal distribution table.
MORE THAN 48:
Again we find the z-score using the formula x-mean/Std.Dev. **or**
(48-48)/3 = 0
Now we find the corresponding z score to 0 using the standard normal distribution table which = .5000 so:
P(x>48)=.5000
You can use the same calculator steps replacing upperbound with 48 to find your answer or check your work.
Hope this helps!!