Each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds. If a household is selected at random, find the probability of its generating a) Between 27 and 31 pounds per month b) Morethan30.2poundspermonth Assume the variable is approximately normally distributed.

We are given that each month, an American household generates an average of 28 pounds of newspaper for garbage or recycling. Assume the standard deviation is 2 pounds and the variable is approximately normally distributed.

Let X = generation of newspaper for garbage or recycling

The z-score probability distribution for normal distribution is given by;

Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ N(0,1)

where, [tex]\mu[/tex] = population average = 28 pounds

[tex]\sigma[/tex] = population standard deviation = 2 pounds

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X.

(a) Probability of household generating between 27 and 31 pounds per month is given by = P(27 pounds < X < 31 pounds) = P(X < 31 pounds) - P(X [tex]\leq[/tex] 27 pounds)

P(X < 31 pounds) = P( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{31-28}{2}[/tex] ) = P(Z < 1.50) = 0.93319

P(X [tex]\leq[/tex] 27 pounds) = P( [tex]\frac{X-\mu}{\sigma}[/tex] [tex]\leq[/tex] [tex]\frac{27-28}{2}[/tex] ) = P(Z [tex]\leq[/tex] -0.50) = 1 - P(Z < 0.50)

= 1 - 0.69146 = 0.30854

{Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.50 and x = 0.50 in the z table which has an area of 0.93319 and 0.30854 respectively.}

Therefore, P(27 pounds < X < 31 pounds) = 0.93319 - 0.30854 = 0.62465

(b) Probability of household generating more than 30.2 pounds per month is given by = P(X > 30.2 pounds)

P(X > 30.2 pounds) = P( [tex]\frac{X-\mu}{\sigma}[/tex] > [tex]\frac{30.2-28}{2}[/tex] ) = P(Z > 1.10) = 1 - P(Z [tex]\leq[/tex] 1.10)

= 1 - 0.8643 = 0.1357

Now, in the z table the P(Z [tex]\leq[/tex] x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 1.10 in the z table which has an area of 0.8643.

raphealnwobi raphealnwobi · Answer 2 · 2021-11-25T13:24:25+01:00

If a household is selected at random, the probability of its generating more than 30.2 pounds per month is 13.57%

Z score is given by:

[tex]z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation\\\\\\Given\ that\ \mu=28,\sigma=2\\\\\\a) For\ x=27:\\\\z=\frac{27-28}{2}=-0.5\\\\For \ x=31:\\\\z=\frac{31-28}{2}=1.5[/tex]

From the normal distribution table, P(27 < x < 31) = P(-0.5 < z < 1.5) = P(z < 1.5) - P(z < -0.5) = 0.9332 - 0.3085 = 62.47%

b) For x > 30.2

[tex]z=\frac{30.2-28}{2} =1.1[/tex]

From the normal distribution table, P(x > 30.2) = P(z > 1.1) = 1 - P(z > 1.1) = 1 - 0.8643 = 13.57%

If a household is selected at random, the probability of its generating more than 30.2 pounds per month is 13.57%

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