Question: Need to show how to calculate the problems for (a) and(b). The gypsy m

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Question: Need to show how to calculate the problems for (a) and(b). The gypsy moth is a serious threat to oak and aspen trees. Astate agriculture department places traps thoughout the state todetect the moths. When traps are checked periodically, the meannumber of moths trapped is only 0.5, but some traps have severalmoths. The distribution of moth counts is discrete and stronglyskewed, with standard deviation 0.7.                                                                                                                       _ (a) What are the mean and standard deviation of the averagenumber of moths x in 50 traps? (b) Use the central limit theorem to find the probability thatthe average number of moths in 50 traps is greater than 0.6. Thank you very much. Question: Need to show how to calculate the problems for (a) and(b). The gypsy moth is a serious threat to oak and aspen trees. Astate agriculture department places traps thoughout the state todetect the moths. When traps are checked periodically, the meannumber of moths trapped is only 0.5, but some traps have severalmoths. The distribution of moth counts is discrete and stronglyskewed, with standard deviation 0.7.                                                                                                                       _ (a) What are the mean and standard deviation of the averagenumber of moths x in 50 traps? (b) Use the central limit theorem to find the probability thatthe average number of moths in 50 traps is greater than 0.6. Thank you very much.

Explanation / Answer

=0.5, =0.7,n=50 According to the central limit theorem, the mean value of a samplewith a large enough sample size is normally distributed. If a sample of size n is drawn from a population with mean and standard deviation , then the sample mean Xbar isnormally distributed with mean and standard deviation/n. Any normally distributed variable X with mean and standarddeviation , X ~ N( , ² ), can be transformed into standardnormal distribution by Z = ( X - ) / , where Z ~ N( 0, 1) ==> The standard normal variable Z has =0 and =1. There are tables of the normal distribution, where you can read thevalue of the standard normal cumulative probability F(z) or(z). (a) Here, according to the central limit theorem, the averagenumber of moths in 50 traps is normally distributed with mean(Xbar)==0.5 and standarddeviation=0=/n=0.7/50=0.098995 (b) z=(0.6-)/0=(0.6-0.5)/0.098995=1.01015 P(Xbar>0.6)=P(z>1.01015)=1-F(1.01015)= 1-0.84379=0.15621

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