x. A pest control company uses a certain pesticide as a perimeter barrier for re

Question

x. A pest control company uses a certain pesticide as a perimeter barrier for residential omers. The pesticide has a mean effectiveness time 37.8 days with a standard deviation of 7.8 days. Suppose that a random sample of 50 homes is chosen and the effectiveness time is measured for each applications of the pesticide. a) What is the distribution for possible samples means? b) Determine the probability that the sample mean obtained will be less than 40 days. Determine the probability that the sampling error from a sample of 50 homes will be les than 3 days o)

Explanation / Answer

a)
The probability distribution of this statistic is caleed sampling distribution.
mean = 37.8 , sd = 7.8 , n =50

b)
P(X <40)
z = (x -mean) /(s/sqrt(n))
z = (40 - 37.8) / (7.8/sqrt(50)
= 1.9944

P(x <40) = P(z <1.9944 ) = 0.9769 by using standard normal table

c) p(x <3)

z = (x -mean) /(s/sqrt(n))
z = (3 - 1.1) / (7.8/sqrt(50)
= 1.7

P(x <40) = P(z <1.7 ) = 0.9554 by using standard normal table

a)
Probability distribution

n = 391 , p = 0.8307

b)
p(pcap > 0.90)
z =( pcap - p) / sqrt(p * (1 -p)/n)
= ( 0.90 - 0.8307) / sqrt(0.8307 * ( 1 -0.8307)/391)
= 3.6540
p(pcap > 0.90) = p(z >3.6540) = 0.0001 by uisng standard normal table

c)
p(pcap < 0.025)
z =( pcap - p) / sqrt(p * (1 -p)/n)
= ( 0.025 - 0.019)/sqrt(0.8307 * ( 1 -0.8307)/391)
= 0.316

p(pcap < 0.025) = p(z <0.316) = 0.624 by uisng standard normal table

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