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 (Round to three decimal places as needed.)

The probability \( P(x) \) for a uniform distribution can be calculated using the formula: \[ P(x \text{ between } a \text{ and } b) = \frac{\text{Length of the interval}}{\text{Total length of distribution}} \] Here, the total length of the distribution is \( b - a = 25 - 5 = 20 \). (a) For \( x \) lying between 6 and 15, the length of the interval is \( 15 - 6 = 9 \). Thus, \[ P(6 < x < 15) = \frac{9}{20} = 0.450 \] So the probability that \( x \) lies between 6 and 15 is \( 0.450 \) (rounded to three decimal places). Now let’s calculate the remaining probabilities: (b) For \( x \) lying between 11 and 24, the length of the interval is \( 24 - 11 = 13 \). The probability is: \[ P(11 < x < 24) = \frac{13}{20} = 0.650 \] (c) For \( x \) lying between 9 and 23, the length of the interval is \( 23 - 9 = 14 \). The probability is: \[ P(9 < x < 23) = \frac{14}{20} = 0.700 \] (d) For \( x \) lying between 7 and 16, the length of the interval is \( 16 - 7 = 9 \). The probability is: \[ P(7 < x < 16) = \frac{9}{20} = 0.450 \] In summary: (a) \( 0.450 \) (b) \( 0.650 \) (c) \( 0.700 \) (d) \( 0.450 \)