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 per trap is only 1.2, but some traps have several moths. The distribution of moth counts in traps is strongly right skewed, with standard deviation 1.4. A random sample of 60 traps has x = 1 and s = 2.4.
Let X = the number of moths in a randomly selectd trap

(a) For the population distribution, what is the ...

...mean? =
...standard deviation? =

(b) For the distribution of the sample data, what is the ...

...mean? =
...standard deviation? =

(c) What shape does the distribution of the sample data probably have?

Exactly NormalApproximately Normal Right skewedLeft skewed



(d) For the sampling distribution of the sample mean with n = 60, what is the ...

...mean? =
...standard deviation? = (Use 3 decimal places)

(e) What is the shape of the sampling distribution of the sample mean?

Left skewedApproximately Normal Right skewedExactly Normal


(f) If instead of a sample size of 60, suppose the sample size were 10 instead. What is the shape of the sampling distribution of the sample mean for samples of size 10?

Approximately normalSomewhat left skewed Exactly NormalSomewhat right skewed


(g) Can we use the Z table to calculate the probability a randomly selected sample of 10 traps has a sample mean less than 1?

No, because the sampling distribution of the sample mean is somewhat right skewedYes, by the Central Limit Theorem, we know the sampling distribution of the sample mean is normal

Answer 1

Answer:

Step-by-step explanation:

Hello!

X: number of gypsy moths in a randomly selected trap.

This variable is strongly right-skewed. with a standard deviation of 1.4 moths/trap.

The mean number is 1.2 moths/trap, but several have more.

a.

The population is the number of moths found in traps places by the agriculture departments.

The population mean μ= 1.2 moths per trap

The population standard deviation δ= 1.4 moths per trap

b.

There was a random sample of 60 traps,

The sample mean obtained is X[bar]= 1

And the sample standard deviation is S= 2.4

c.

As the text says, this variable is strongly right-skewed, if it is so, then you would expect that the data obtained from the population will also be right-skewed.

d. and e.

Because you have a sample size of 60, you can apply the Central Limit Theorem and approximate the distribution of the sampling mean to normal:

X[bar]≈N(μ;σ²/n)

The mean of the distribution is μ= 1.2

And the standard deviation is σ/√n= 1.4/50= 0.028

f. and g.

Normally the distribution of the sample mean has the same shape of the distribution of the original study variable. If the sample size is large enough, as a rule, a sample of size greater than or equal to 30 is considered sufficient you can apply the theorem and approximate the distribution of the sample mean to normal.

You have a sample size of n=10 so it is most likely that the sample mean will have a right-skewed distribution as the study variable. The sample size is too small to use the Central Limit Theorem, that is why you cannot use the Z table to calculate the asked probability.

I hope it helps!