Q.7 - The Practice of Statistics for AP

Question

The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths trapped is only $0.5,$ but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation$0.7$.

 (a) What are the mean and standard deviation of the average number of moths $x$ in$50$ traps?

 (b) Use the central limit theorem to help you find the probability that the average number of moths in $50$ traps is greater than$0.6$.

Step-by-Step Solution

Answer

(a)Mean$=0.5$

Standard deviation$=0.0990$

(b)Probability is $0.1562$

Part(a) Step 1: Given information

The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths trapped is only $0.5$, but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation $0.7$. 

Part(a) Step 2: Explanation

Population mean$\left(\mu \right)=0.50$

Population standard deviation$\left(\sigma \right)=0.7$

Sample size $\left(n\right)=50$

The mean and standard deviation can be calculated as:

${\mu }_{\overline{x}}=\mu =0.5$

$$\begin{gathered} {\sigma }_{\overline{x}}=\frac{\sigma }{\sqrt{n}} \\ =\frac{0.7}{\sqrt{50}} \\ =0.0990 \end{gathered}$$

The required mean and standard deviation are $0.5$and $0.0990$ respectively

Part(b) Step 1: Given information

The gypsy moth is a serious threat to oak and aspen trees. A state agriculture department places traps throughout the state to detect the moths. When traps are checked periodically, the mean number of moths trapped is only $0.5$, but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation $0.7.$

Part(b) Step 2: Explanation

The probability that average number of months is greater than $0.6$ can be calculated as:

$$\begin{gathered} P(\overline{x}\ge 0.6)=P\left(Z\ge \frac{0.6-0.5}{\frac{0.7}{\sqrt{30}}}\right) \\ =P(Z\ge 1.01) \\ =0.1562 \end{gathered}$$

Thus, the probability is $0.1562$ .