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The distribution is best described as: \( \square \) Uniform The correct notation for this density curve is: \( \square \) \( U(16,24) \) Compute the following probabilities. State answers as decimals. \[ \begin{array}{l} P(X \leq 18)=\text { type your answer... } \\ P(X>19)=\text { type your answer... } \end{array} \] Calculate the following: \[ Q_{1}=18 \] BMedian = \( \square \) 20 \[ Q_{3}= \] \[ \mathrm{IQR}= \] \( \square \) \( \square \) \[ \mu= \] \( \square \) \[ \sigma= \] \( \square \)

Ask by Wagner Gough. in the United States
Feb 25,2025

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Distribution: Uniform Notation: U(16, 24) - P(X ≤ 18) = 0.25 - P(X > 19) = 0.625 - Q₁ = 18 - Median = 20 - Q₃ = 22 - IQR = 4 - μ = 20 - σ ≈ 2.31

Solución

We are given a Uniform distribution on the interval [16, 24], which we denote as U(16, 24). 1. Probabilities  a. P(X ≤ 18)   For a uniform distribution, the probability is proportional to the length. Thus:    P(X ≤ 18) = (18 – 16) / (24 – 16) = 2/8 = 0.25  b. P(X > 19)   Here, compute the length from 19 to 24:    P(X > 19) = (24 – 19) / (24 – 16) = 5/8 = 0.625 2. Quantiles and Related Measures  a. Q1 is given as 18.  b. The median (50th percentile) is computed by:    Median = (16 + 24) / 2 = 40/2 = 20  c. Q3 is the 75th percentile:    Q3 = 16 + 0.75*(24 – 16) = 16 + 0.75*8 = 16 + 6 = 22  d. IQR (Interquartile Range) = Q3 – Q1 = 22 – 18 = 4 3. Mean and Standard Deviation  a. The mean (μ) for a uniform distribution is:    μ = (16 + 24) / 2 = 20  b. The standard deviation (σ) is given by:    σ = (b – a) / √12 = 8 / √12 = 8 / (2√3) = 4 / √3    Approximating √3 ≈ 1.732, we have:    σ ≈ 4 / 1.732 ≈ 2.31 Summary of answers: • Distribution: Uniform • Correct notation: U(16, 24) • P(X ≤ 18) = 0.25 • P(X > 19) = 0.625 • Q1 = 18 • Median = 20 • Q3 = 22 • IQR = 4 • μ = 20 • σ ≈ 2.31

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For a uniform distribution \( U(16,24) \), the probability density function is constant between the parameters 16 and 24. Therefore, the length of the interval is \( 24 - 16 = 8 \). To find \( P(X \leq 18) \): \[ P(X \leq 18) = \frac{18 - 16}{24 - 16} = \frac{2}{8} = 0.25. \] To find \( P(X > 19) \): \[ P(X > 19) = \frac{24 - 19}{24 - 16} = \frac{5}{8} = 0.625. \] Now for the quartiles and the mean and standard deviation. \[ Q_{1} = 18, \] \[ BMedian = 20, \] \[ Q_{3} = 22, \] \[ \mathrm{IQR} = Q_{3} - Q_{1} = 22 - 18 = 4. \] The mean \( \mu \) of a uniform distribution is calculated as: \[ \mu = \frac{a + b}{2} = \frac{16 + 24}{2} = 20. \] The standard deviation \( \sigma \) is given by: \[ \sigma = \sqrt{\frac{(b - a)^2}{12}} = \sqrt{\frac{(24 - 16)^2}{12}} = \sqrt{\frac{64}{12}} \approx 2.31. \] So, summarizing: \[ P(X \leq 18) = 0.25, \] \[ P(X > 19) = 0.625, \] \[ Q_{3} = 22, \] \[ \mathrm{IQR} = 4, \] \[ \mu = 20, \] \[ \sigma \approx 2.31. \]

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