View Full Version : Probability question
asusenna
Oct 21, 2010, 03:08 PM
Subway passengers spend an average of $5 a day on subway fares. The standard deviation of the expenditure is $3. A simple random sample of 36 passengers is taken. What is the probability that the sample mean will be between $4 and $7?
Unknown008
Oct 21, 2010, 10:46 PM
Are you sure this is the correct question? I'm a little puzzled...
The mean is given by \bar{x} = np
The standard deviation by Sd(x) = \sqrt{npq}
From this, we get q = 1.8 which is not possible... Also, we usually have cases up to three standard deviations from the mean, meaning that there will be passengers paying "-$4"? Either the standard deviation is too large, or the mean too small.
Anyway, if that is indeed the correct question, it really isn't very realistic.
We'll use a normal distribution.
Let X be the price a passenger pays.
X ~ N(5, 3^2)
P(4 < X < 7) = P(-0.333 < z < 0.667) = P(z < 0.667) - P(z < -0.333)
Which gives a probability of 0.3777
asusenna
Oct 22, 2010, 02:38 PM
Thank you for your help. However this is a wrong solution.
Correct solution is: we insert in TI-83 the smallest value, the largest value, mean and standard deviation Q/√N. Put normalcdf on calculator and insert 4, 7, 5, 0.5 = 0.9772
Thank you
Unknown008
Oct 22, 2010, 11:30 PM
You gave the standard deviation as $3 and not $0.5... :rolleyes:
Or that could be a distribution I haven't learned.
galactus
Oct 23, 2010, 05:20 AM
To find the probability that a sample mean will fall in a given interval, we can use
x=\frac{\text{value}-\text{mean}}{\text{standad deviation}}
z=\frac{(x-{\mu})\sqrt{n}}{\sigma}
In this case,
\frac{(4-5)6}{3}=-2
The z-score is .02275
\frac{(7-5)6}{3}=4
The z-score is .9999683
Subtract to the find the probability it will fall between 4 and 7 and we get
.977218
There is almost a 98% probability the sample mean will be between 4 and 7.
Unknown008
Oct 23, 2010, 06:31 AM
Hm... definitely unknown to me ;)
Yes, I never worked with a probability sample mean being within a range, but only a certain value.
I had defined my variable as being for one passenger. I wonder if this would still hold... Hm... we multiply the z value otherwise obtained by the square root of the sample size and take the probabilities... I'll try to remember that.