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?

Are you sure this is the correct question? I'm a little puzzled...

The mean is given by
The standard deviation by

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

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

You gave the standard deviation as $3 and not $0.5... :rolleyes:

Or that could be a distribution I haven't learned.

To find the probability that a sample mean will fall in a given interval, we can use





In this case,



The z-score is .02275



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.

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.