Parcels of printed matter being posted from the office are always weighted. The weights have been found to be normally distributed with a mean of 8.35kg and a standard deviation of 2.25kg.Find the proportion of parcels whose weights are (i) less than 10kg (Give answer correct to 3 decimal places) (ii) between 6.5kg and 9kg. (Give answer correct to 3 decimal places) (iii) What weight would a parcel need to be to be in the top 15% of weights? (Give answer correct to 2 decimal places)
Question
Parcels of printed matter being posted from the office are always weighted. The weights have been found to be normally distributed with a mean of 8.35kg and a standard deviation of 2.25kg.Find the proportion of parcels whose weights are (i) less than 10kg (Give answer correct to 3 decimal places) (ii) between 6.5kg and 9kg. (Give answer correct to 3 decimal places) (iii) What weight would a parcel need to be to be in the top 15% of weights? (Give answer correct to 2 decimal places)
Solution
To solve this problem, we will use the properties of the normal distribution and the concept of z-scores.
(i) To find the proportion of parcels that weigh less than 10kg, we first need to convert the weight of 10kg to a z-score. The z-score is calculated as follows:
Z = (X - μ) / σ
where X is the value we are interested in (10kg), μ is the mean (8.35kg), and σ is the standard deviation (2.25kg).
Z = (10 - 8.35) / 2.25 = 0.733
We then look up this z-score in a standard normal distribution table or use a calculator with a normal distribution function to find the proportion of values less than this. The proportion is 0.732.
(ii) To find the proportion of parcels that weigh between 6.5kg and 9kg, we first convert these weights to z-scores:
Z1 = (6.5 - 8.35) / 2.25 = -0.822 Z2 = (9 - 8.35) / 2.25 = 0.289
We then find the proportion of values less than these z-scores and subtract the smaller from the larger:
P(Z < 0.289) - P(Z < -0.822) = 0.614 - 0.206 = 0.408
(iii) To find the weight that a parcel would need to be to be in the top 15% of weights, we first find the z-score that corresponds to the 85th percentile (since 100% - 15% = 85%). This z-score is approximately 1.036. We then convert this z-score back to a weight:
X = μ + Zσ = 8.35 + 1.036(2.25) = 10.37kg
So a parcel would need to weigh more than 10.37kg to be in the top 15% of weights.
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