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 . 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 (Round to three decimal places as needed.)

The uniform distribution is a continuous probability distribution where all outcomes are equally likely within a defined range. The probability density function (PDF) for a uniform distribution is given by \( f(x) = \frac{1}{b - a} \) for \( a \leq x \leq b \) and 0 otherwise. For your distribution from \( a = 5 \) to \( b = 25 \), the total width of the interval is \( 20 \). The formula for the probability of \( x \) lying between two values \( c \) and \( d \) is given by: \[ P(c \leq x \leq d) = \frac{d - c}{b - a} \] In (b), to find the probability that \( x \) lies between 11 and 24: \[ P(11 \leq x \leq 24) = \frac{24 - 11}{25 - 5} = \frac{13}{20} = 0.650 \] For (c) between 9 and 23: \[ P(9 \leq x \leq 23) = \frac{23 - 9}{25 - 5} = \frac{14}{20} = 0.700 \] For (d) between 7 and 16: \[ P(7 \leq x \leq 16) = \frac{16 - 7}{25 - 5} = \frac{9}{20} = 0.450 \] So in summary: (b) The probability that \( x \) lies between 11 and 24 is \( 0.650 \). (c) The probability that \( x \) lies between 9 and 23 is \( 0.700 \). (d) The probability that \( x \) lies between 7 and 16 is \( 0.450 \).