I was tutoring a statistics student who had this problem that I cant figure out.
Sacks of grain are normally distributed with mean a 100 pounds and standard deviation 2 pounds. A wagon carries 9 sacks of grain what is the probability that the wagonload weighs more than 910 pounds?
- a. between .045 and .049
- b. between .101 and .105
- c. between .287 and .291
- d. between .019 and .023
- e. between .151 and .155
My inclination is to use a standard z table problem:
$ Z = \frac{\overline{X} - \mu}{\sigma/\sqrt n}$
$ Z = \frac{101.1 - 100}{2/\sqrt 9}$
Which yields a Z value of approximately 1.65 leading to a probability 95% so an answer of 5%. My question is why is a range of probabilities needed and how is it computed and what is the correct answer?
You've done everything right, except that you should've possibly taken a few more decimals of $\frac{910}{9}$. Actually:
$$Z=\frac{\frac{10}{9}}{\frac{2}{3}}=\frac{5}{3}=1.666\ldots$$
Now, putting that into Z-tables, which, in my case, expect the argument to two decimal places... It is between the values for $1.66$ and $1.67$, which are $0.9515$ and $0.9525$, so the requested probability is between $0.0475$ and $0.0485$, which fits the answer (a).
The ranges are, I guess, given so that the students can use a range of Z tables (or possibly calculators) and still arrive at the same conclusion.