MATH SOLVE

10 months ago

Q:
# 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 0.6, but some traps have several moths. The distribution of moth counts is discrete and strongly skewed, with standard deviation 0.4.a. What is the mean (±0.1) of the average number of moths x¯¯¯ (x bar) in 30 traps?I got 18 by multiplying 30 times 0.6, I'm not confident 100% in thisb. And the standard deviation? (±0.001) I got 0.073 by doing 0.4/ sqrt of 30...not confidentc. Use the central limit theorem to find the probability (±0.01) that the average number of moths in 30 traps is greater than 0.7: I got 1.37 by doing 0.7-0.6 / 0.073 (0.4/sqrt of 30)are these answers correct?

Accepted Solution

A:

Part A:

From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.

Part B:

The standard deviation is given by the population deviation divided by the square root of the sample size.

[tex]standard \ deviation= \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{30}} = \frac{0.4}{5.477} =0.073[/tex]

Part C:

The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by

[tex]P(X\ \textgreater \ x)=1-P(X\ \textless \ x) \\ \\ =1-P\left(z\ \textless \ \frac{x-\mu}{\sigma/\sqrt{n}} \right)[/tex]

Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that the average number of moths in 30 traps is greater than 0.7 given by:

[tex]P(X\ \textgreater \ 0.7)=1-P(X\ \textless \ 0.7) \\ \\ =1-P\left(z\ \textless \ \frac{0.7-0.6}{0.073} \right)=1-P(z\ \textless \ 1.369) \\ \\ =1-0.91455=\bold{0.0855}[/tex]

From the central limit theorem, since the number of samples is large enough (up to 30), the mean of the the mean of the average number of moths in 30 traps is 0.6.

Part B:

The standard deviation is given by the population deviation divided by the square root of the sample size.

[tex]standard \ deviation= \frac{\sigma}{\sqrt{n}} = \frac{0.4}{\sqrt{30}} = \frac{0.4}{5.477} =0.073[/tex]

Part C:

The probability that an approximately normally distributed data with a mean, μ, and the standard deviation, σ, with a sample size of n is greater than a number, x, given by

[tex]P(X\ \textgreater \ x)=1-P(X\ \textless \ x) \\ \\ =1-P\left(z\ \textless \ \frac{x-\mu}{\sigma/\sqrt{n}} \right)[/tex]

Thus, given that the mean is 0.6 and the standard deviation is 0.4, the probability that the average number of moths in 30 traps is greater than 0.7 given by:

[tex]P(X\ \textgreater \ 0.7)=1-P(X\ \textless \ 0.7) \\ \\ =1-P\left(z\ \textless \ \frac{0.7-0.6}{0.073} \right)=1-P(z\ \textless \ 1.369) \\ \\ =1-0.91455=\bold{0.0855}[/tex]