.S Answer the following assuming a standard normal distribution a) Ifthe value i

Question

.S Answer the following assuming a standard normal distribution a) Ifthe value is 1.96, find the probability between 0 and the r-vae b) Find the z-value of area under curve between 0 and 0.2133 c) Each month, an American household generates an average of 30 pounds of newspaper for recycling. Assume the standard deviation is 3 pounds. Ifa household is selected at random, find the probability of generating between 27 and 34.5 pounds month. Also, find the probability of generating more than 37.5 pounds per month. 3S-30

Explanation / Answer

Solution:

Probability between z value 0 to 1.96

P(0<Z<1.96) = p(z<1.96)- p(z<0)

From z table we found probability of z values

= 0.9750-0.5 = 0.4750 or 47.50%

Solution(b)

if p-value is 0 than from z table we found that z value is -4.75

If p-value is 0.2133 than from z table we found z value is -0.79502

So z value is (-4.75<Z<-0.79502

Solution(c)

Mean =30

SD =3

P(27<xbar<34.5)=p(xbar<34.5)-p(xbar<27)

Z=( 27-30)/3 =-1

Z = (34.5-30)/3 = 1.5

So P(27<Xbar<34.5) = 0.9332-0.8413 = 0.0919 or 9.19%

P(xbar>37.5)

Z = (37.5-30)/3 = 7.5/3 = 2.5

P(xbar>37.5) =1-0.9938 = 0.0062 or 0.62%

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