SOLUTION: Assume that male weight is distributed Uniformly from 140 to 260 lbs . a. What is the mean and standard deviation of the distribution for male weight? b. What is the prob

Question 1177194: Assume that male weight is distributed Uniformly from 140 to 260 lbs .
a. What is the mean and standard deviation of the distribution for male weight?

b. What is the probability that a randomly chosen male weighs less than 175?

c. What is the probability that a randomly chosen male weighs more than 215?

d. What is the probability that a randomly chosen male weighs between 180 and 190 lbs?

e. Assume that you sample 10 males and find their average weight. What is the mean and standard deviation of the distribution for average weight of 10 males?

f. Assuming that 10 is a “large” sample, what is the probability that the average weight for the 10 males is between 180 and 190 lbs?

Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's break down this problem step-by-step.
**Given:**
* Male weight is uniformly distributed from 140 lbs to 260 lbs.
* a = 140 (lower bound)
* b = 260 (upper bound)
**a. Mean and Standard Deviation:**
* **Mean (μ):**
* μ = (a + b) / 2
* μ = (140 + 260) / 2 = 400 / 2 = 200 lbs
* **Standard Deviation (σ):**
* σ = √((b - a)² / 12)
* σ = √((260 - 140)² / 12)
* σ = √(120² / 12) = √(14400 / 12) = √1200 ≈ 34.64 lbs
**b. Probability (Weight < 175 lbs):**
* P(X < 175) = (175 - a) / (b - a)
* P(X < 175) = (175 - 140) / (260 - 140) = 35 / 120 ≈ 0.2917
**c. Probability (Weight > 215 lbs):**
* P(X > 215) = (b - 215) / (b - a)
* P(X > 215) = (260 - 215) / (260 - 140) = 45 / 120 ≈ 0.375
**d. Probability (180 lbs < Weight < 190 lbs):**
* P(180 < X < 190) = (190 - 180) / (b - a)
* P(180 < X < 190) = 10 / 120 ≈ 0.0833
**e. Mean and Standard Deviation of Sample Mean (n = 10):**
* **Mean (μₓ̄):**
* μₓ̄ = μ = 200 lbs (The mean of the sample mean is the same as the population mean)
* **Standard Deviation (σₓ̄):**
* σₓ̄ = σ / √n
* σₓ̄ = 34.64 / √10 ≈ 10.95 lbs
**f. Probability (180 lbs < Sample Mean < 190 lbs, n = 10):**
Since n=10 is assumed to be large, we can apply the central limit theorem and assume the sample mean is normally distributed.
We need to calculate the z-scores for 180 lbs and 190 lbs.
* z1 = (180-200)/10.95 = -1.8265
* z2 = (190-200)/10.95 = -0.9132
Now we need to find the probability P(-1.8265 < Z < -0.9132) using a standard normal distribution table or calculator.
* P(Z < -0.9132) = 0.1803
* P(Z < -1.8265) = 0.0339
* P(-1.8265 < Z < -0.9132) = P(Z < -0.9132) - P(Z < -1.8265) = 0.1803 - 0.0339 = 0.1464
**Summary of Answers:**
* **a.** Mean = 200 lbs, Standard Deviation ≈ 34.64 lbs
* **b.** Probability ≈ 0.2917
* **c.** Probability ≈ 0.375
* **d.** Probability ≈ 0.0833
* **e.** Mean = 200 lbs, Standard Deviation ≈ 10.95 lbs
* **f.** Probability ≈ 0.1464

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