The mean amount of time it takes a kidney stone to pass is 13 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.

X ~ Norm ( 13 , 25 )

P ( X > 17 ) = 0.2119

16.37 days

Step-by-step explanation:

Solution:-

- We are given a distribution for the amount of time for a kidney stone to pass.

- The distribution is parameterize by the mean time taken ( u ) and the standard deviation ( s ) as follows:

                                 u = 13 days

                                 s = 5 days

- Here, we will define a random variable X: The time taken by a kidney stone to pass to be normally distributed with parameters ( u ) and ( s ). We express the distribution in the notation form as follows:

                              X ~ Norm ( u , s^2 )

                              X ~ Norm ( 13 , 25 )

- We are to determine that a randomly selected individual takes more than 17 days for the stone to pass through.

- We will first standardize the limiting value for the required probability by computing the Z-score as follows:

- We will use the standard normal table to determine the probability of kidney stone passing in less than 17 days ( Z = 0.8 ); hence, we have:

                           P ( X < 17 ) = P ( Z < 0.8 )

                           P ( X < 17 ) = 0.7881

- To compute the probability of an individual taking more than 17 days would be " total probability - P ( X < 17 )  as follows" . Where the total probability of any distribution is always equal to 1.

                        P ( X > 17 ) = 1 - P ( X < 17 )

                        P ( X > 17 ) = 1 - 0.7781

                        P ( X > 17 ) = 0.2119

- Nest we are to determine the amount of days it would take for an individual to lie in the upper quarter of the spectrum. We can interpret this by looking at the limiting value corresponding to the P ( X > x ) = 0.25.

- The upper quartile of any distribution amounts to probabilities: " > x = 0.25 " or " < x = 0.75 ".

- We will use the standard normal table for ( Z-score ) and look-up the Z-score value corresponding to P ( Z < a ) = 0.75 as follows:

                       P ( Z < a ) = 0.75

                       a = 0.674

- Now we will use the standardizing formula used in previous part and compute the value of "x" associated with the limiting Z-score value:

It would should take more than 16.37 days for an individual if he is to lie in the upper quartile of the defined distribution.

hiking down a winding trail.

step-by-step explanation: