Questions: Homework - Ch. 6: Normal Dist Assignment - Ch. 6: Normal Distributions - Calculations (Thomas) Score: 7.5/100 Answered: 1/9 Question 3 The mean amount of time it takes a kidney stone to pass is 14 days and the standard deviation is 5 days. Suppose that one individual is randomly chosen. Let X= time to pass the kidney stone. Round all answers to 4 decimal places where possible. a. What is the distribution of X ? X-N b. Find the probability that a randomly selected person with a kidney stone will take longer than 15 days to pass it. c. Find the minimum number for the upper quarter of the time to pass a kidney stone. days.

What is the distribution of \( X \)?

Distribution of \( X \)

The random variable \( X \), representing the time to pass a kidney stone, follows a normal distribution with a mean of \( 14 \) days and a standard deviation of \( 5 \) days. Thus, we can express this as: \[ X \sim N(14, 5^2) \]

\(\boxed{X \sim N(14, 5^2)}\)

Find the probability that a randomly selected person with a kidney stone will take longer than \( 15 \) days to pass it.

Calculate the cumulative distribution function (CDF) for \( X \) at \( x = 15 \)

The probability that a randomly selected person will take longer than \( 15 \) days is given by: \[ P(X > 15) = 1 - P(X \leq 15) = 1 - CDF(15) \] Calculating this gives: \[ P(X > 15) \approx 0.4207 \]

\(\boxed{P(X > 15) \approx 0.4207}\)

Find the minimum number for the upper quarter of the time to pass a kidney stone.

Determine the 75th percentile (upper quarter) of the distribution

The minimum time for the upper quarter can be found using the inverse cumulative distribution function (inverse CDF) at \( p = 0.75 \): \[ X_{0.75} \approx 17.3724 \text{ days} \]

\(\boxed{X_{0.75} \approx 17.3724 \text{ days}}\)

The distribution of \( X \) is \( X \sim N(14, 5^2) \).

The probability that it will take longer than \( 15 \) days is approximately \( 0.4207 \).

The minimum number for the upper quarter of the time is approximately \( 17.3724 \) days.