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 \[ Q_{3}= \] \[ \mathrm{IQR}= \] \( \square \) \( \square \) \[ \mu= \] \( \square \) \[ \sigma= \] \( \square \)
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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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