# sampling distribution with no continuity correction

Assume that the distribution of the height of a plant is normally distributed with a mean height of $\mu = 83.4$ inches and standard deviation of $\sigma = 9.1$ inches. Find:

1. if three such plants are chosen at random find $\Pr[ \bar y > 90 {\rm \; inches}]$.
2. if two such plants are chosen at random, then what is the probability that one plant will be $90$ or more inches while the other plant will be less than $90$ inches tall? (no continuity correction)
3. if three such plants are chosen at random, then what is the probability that two out of three plants will be $90$ or more inches tall? (no continuity correction)

For #1 I think it should be like this : $\frac{90-83.4}{9.1/\sqrt{4}}$ which would be $0.3409$ but I'm not sure about #2 and #3.

1) The random variable $\bar{Y}$ has mean $83.4$ and standard deviation $\frac{9.1}{\sqrt{3}}\approx 5.254$. Thus $$\Pr(\bar{Y}\gt 90)=\Pr\left(Z\gt\frac{90-83.4}{5.254}\right),$$ where $Z$ is standard normal. You can find this by using tables of the standard normal, or software.

2) First we find the probability that an individual plant is $\gt 90$. This is the probability that a normal with mean $83.4$ and standard deviation $9.1$ is greater than $90$. This is the probability that $Z\gt \frac{90-83.4}{9.1}$. You can find this number using tables of the standard normal. Call the number $p$.

Our probability is then probability that the first is $\gt 90$, and the second isn't, plus the probability the first isn't, but the second is. This is $p(1-p)+(1-p)p=2p(1-p)$.

Or else using the Binomial distribution, we find that the probability is $\binom{2}{1}p(1-p)$.

3) We interpret this as meaning that exactly $2$ are $\gt 90$. This can happen in $3$ ways: (i) First is $\gt 90$, second is, third isn't; (ii) First is, second isn't, third is; (iii) First isn't, each of the others is.

Each of these turns out to have probability $p^2(1-p)$, for a total of $3p^2(1-p)$.

Or else use the Binomial Distribution, to conclude answer is $\binom{3}{2}p^2(1-p)$.

• for #1 shouldn't we find the value from the z table then subtract 1 from it ? – kingjames Feb 17 '14 at 7:08
• Yes, I left that part to you, on the (correct) assumption that you knew how to handle tables of the standard normal. Same with finding $p$ in the second problem. – André Nicolas Feb 17 '14 at 7:10