The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.0 minutes and standard deviation 1.3 minutes. Suppose that a random sample of customers is observed. Find the probability that the average time waiting in line for these customers is

P ( 5 < X < 10 ) = 1

Step-by-step explanation:

Given:-

- Sample size n = 49

- The sample mean u = 8.0 mins

- The sample standard deviation s = 1.3 mins

Find:-

Find the probability that the average time waiting in line for these customers is between 5 and 10 minutes.

Solution:-

- We will assume that the random variable follows a normal distribution with, then its given that the sample also exhibits normality. The population distribution can be expressed as:

                                   X ~ N ( u , s /√n )

Where

                            s /√n = 1.3 / √49 = 0.2143

- The required probability is P ( 5 < X < 10 ) minutes. The standardized values are:

                        P ( 5 < X < 10 ) = P (    (5 - 8) / 0.2143 <  Z  <  (10-8) / 0.2143   )

                                                 = P ( -14.93 < Z < 8.4 )

- Using standard Z-table we have:

                        P ( 5 < X < 10 ) = P ( -14.93 < Z < 8.4 ) = 1        

idk

step-by-step explanation:

payback

$75.00

step-by-step explanation:

i = prt

i = 1,5000 * 0.025 * 2

i = 75