Answer:
A) 1
B) 1
C) 0
Step-by-step explanation:
The amount of time that a customer spends waiting at an airport check-in counter is a random variable with mean 8.2 minutes and standard deviation 1.5 minutes. Suppose that a random sample of n = 49 customers is observed. Find the probability that the average time waiting in line for these customers is A) Less than 10 minutes B) Between 5 and 10 minutes C) Less than 6 minutes
We solve this question using the z score formula
z = (x-μ)/σ/√n, where
x is the raw score
μ is the population mean
σ is the population standard deviation.
n = number of random samples
A) Less than 10 minutes
x < 10
z = 10 - 8.2/ 1.5 / √49
z = 8.4
P-value from Z-Table:
P(x<10) = 1
B) Between 5 and 10 minutes
For x = 5 minutes
z = 5 - 8.2/ 1.5 / √49
z = -14.93333
P-value from Z-Table:
P(x = 5) = 0
For x = 10 minutes
z = 10 - 8.2/ 1.5 / √49
z = 8.4
P-value from Z-Table:
P(x = 10) = 1
The probability that the average time waiting in line for these customers is between 5 and 10 minutes
P(x = 10) - P(x = 5)
= 1 - 0
= 1
C) Less than 6 minutes
x < 6
z = 6 - 8.2/ 1.5 / √49
z = -10.26667
P-value from Z-Table:
P(x<6) = 0
Well what’s the table of values ?
1/4 * 496 = .25 * 496 = 124
124 flights left the airport
The value in the tenths place would be 7
The amount of time it would will it take him to achieve his goal is B. $5,136; approximately 24 months.
<h3>What are Savings?</h3>
This refers to the emergency funds that are set aside for situations that were not initially budgeted for.
Hence, because Sammy makes about $2,362/month
Rent, car insurance $1,292
Basic groceries, telephone $420
Snack items $260
Annual Savings $130
Annual emergency savings $210
5 years of vacation savings $50
Total= $2,362
Using Sammy's method of putting these budgets into separate checking accounts, the amount which Sammy needs to save to have x3 of his living expenses and the length of time it would take is $5,136; approximately 24 months.
Read more about savings here:
brainly.com/question/25787382
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