Consider a uniform distribution from \( \mathrm{a}=5 \) to \( \mathrm{b}=25 \) (a) Find the probability that x lies between 6 and 15 (b) Find the probability that \( x \) lies between 11 and 24 (c) Find the probability that \( x \) lies between 9 and 23 (d) Find the probability that \( x \) lies between 7 and 16 . E Click the icon to see the definition of the uniform distribution. (a) The probability that \( x \) lies between 6 and 15 is 0.450 . (Round to three decimal places as needed.) (b) The probability that \( x \) lies between 11 and 24 is 0.650 (Round to three decimal places as needed.) (c) The probability that \( x \) lies between 9 and 23 is 0.700 (Round to three decimal places as needed.) (d) The probability that \( x \) lies between 7 and 16 is \( \square \). (Round to three decimal places as needed.)
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To find the probabilities for the uniform distribution from \( a = 5 \) to \( b = 25 \), we can use the formula for the probability of an interval \( (x_1, x_2) \): \[ P(x_1 \leq X \leq x_2) = \frac{x_2 - x_1}{b - a} \] where \( b - a \) is the length of the entire distribution. **(a)** You already calculated \( P(6 \leq x \leq 15) \) as 0.450, which is correct! **(b)** For \( P(11 \leq x \leq 24) \): \[ P(11 \leq x \leq 24) = \frac{24 - 11}{25 - 5} = \frac{13}{20} = 0.650 \] **(c)** For \( P(9 \leq x \leq 23) \): \[ P(9 \leq x \leq 23) = \frac{23 - 9}{25 - 5} = \frac{14}{20} = 0.700 \] **(d)** For \( P(7 \leq x \leq 16) \): \[ P(7 \leq x \leq 16) = \frac{16 - 7}{25 - 5} = \frac{9}{20} = 0.450 \] So, the probability that \( x \) lies between 7 and 16 is **0.450** (rounded to three decimal places).
