9. 11 points 0/3 Submissions Used My Notes The weights of a certain type of stee

Question

9. 11 points 0/3 Submissions Used My Notes The weights of a certain type of steer can be described by a Normal model with a mean weight of 1100 pounds and a standard deviation of 81 pounds. (Use 4 decimals for all questions below.) (a) what weight is required for a steer to be among the heaviest 10% of steers? (b) what weight is required for a steer to be among the lightest 25% of steers? (c) What is the IQR of these weights? 10. -14 points 0/3 Submissions Used My Notes The monthly returns for a financial advisory service can be modeled by a Normal distribution with a mean of $172 and standard deviation of $55, per $10,000 invested. Find the following boundaries (use 4 decimals for all answers): (a) the highest 10% of monthly returns (b) the lowest 10% of monthly returns: (c) the highest 15% of monthly returns: (d) the middle 60% of monthly returns and (Enter the lower value first.)

Explanation / Answer

9) Mean = 1100 pounds

Standard deviation = 81 pounds

P(X < A) = P(Z < (A - mean)/standard deviation)

a) Let W denote the weight of heaviest 10%

P(X < W) = 1 - 0.10 = 0.9

P(Z < (W - 1100)/81) = 0.9

(W - 1100)/81 = 1.28 (taking value of Z corresponding to 0.9 from standard normal distribution table)

W = 1203.68 pounds

b) Let V denote the weight of lightest 25%

P(X < V) = 0.25

P(Z < (V - 1100)/81) = 0.25

(V - 1100)/81 = -0.67 (taking value of Z corresponding to 0.25 from standard normal distribution table)

V = 1045.73 pounds

c) Inter Quartile Range, IQR is the difference between 75th percentile and 25th percentile

Weight between 25th percentile and 50th percentile = 1100 - 1045.73

= 54.27

IQR = 54.27x2

= 108.54

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