Question 1177194
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