Consider a uniform distribution from \( a=5 \) to \( b=25 \). (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 . 5 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 andod 23 is (Round to three decimal places as needed.)

To find the probabilities for the uniform distribution between \( a = 5 \) and \( b = 25 \), we'll use the formula for the probability \( P(c \leq x \leq d) \) in a uniform distribution: \[ P(c \leq x \leq d) = \frac{d - c}{b - a} \] ### (c) Probability that \( x \) lies between 9 and 23 Using the formula: \[ P(9 \leq x \leq 23) = \frac{23 - 9}{25 - 5} = \frac{14}{20} = 0.700 \] **Answer:** **(c)** The probability that \( x \) lies between 9 and 23 is **0.700**. (Rounded to three decimal places as needed.) ### Summary of All Parts For completeness, here's a summary of all the parts: - **(a)** \( P(6 \leq x \leq 15) = \frac{15 - 6}{25 - 5} = \frac{9}{20} = 0.450 \) - **(b)** \( P(11 \leq x \leq 24) = \frac{24 - 11}{25 - 5} = \frac{13}{20} = 0.650 \) - **(c)** \( P(9 \leq x \leq 23) = \frac{14}{20} = 0.700 \) - **(d)** \( P(7 \leq x \leq 16) = \frac{16 - 7}{25 - 5} = \frac{9}{20} = 0.450 \) Feel free to ask if you need further assistance with part (d) or any other questions!