Question

The weight distribution of parcels sent in a certain manner is normal with mean value 12lb and standard deviation 3.5lb.

a. Find the probability that a randomly selected parcel weighs between 9lb and 17lb

b. Find the probability that a randomly selected parcel's weight is within 1.5 standard deviations of the mean value

c. Find the 33rd percentile of this weight distribution

Answer 1

Given = 12 lbs and = 3.5 lbs

To find the probability we need to find the z score. Z = (X - ) /

(a) P(9 < X < 17) = P(X < 17) - P(X < 9)

For P( X < 17)

Z = (17 – 12)/3.5 = 1.43

The probability for P(X < 17) from the normal distribution tables is = 0.9236

For P( X < 9)

Z = (9 – 12)/3.5 = -0.86

The probability for P(X < 9) from the normal distribution tables is = 0.1949

Therefore the required probability is 0.9236 – 0.1949 = 0.7287

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(b) Within 1.5 standard deviation of the mean value = 1.5 * 3.5 = 5.25

Therefore P(12 - 5.25 < X < 12 + 5.25) = P(6.75 < X < 17.25) = P(X < 17.25) - P(X < 6.75)

For P( X < 17.25)

Z = (17.25 – 12)/3.5 = 1.5

The probability for P(X < 17.5) from the normal distribution tables is = 0.9332

For P( X < 6.75)

Z = (6.75 – 12)/3.5 = -1.5

The probability for P(X < 6.75) from the normal distribution tables is = 0.0668

Therefore the required probability is 0.9332 – 0.0668 = 0.8664

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(c) The 33rd percentile means

P(X < x) = 0.33

The z value for p = 0.33 is -0.44

Therefore -0.44 = (X - 12) / 3.5.

Solving for X, we get X = (-0.44 * 3.5) + 12 = 11.69

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