Answer : 0.0129
Step-by-step explanation:
Given : Based on FAA estimates the average age of the fleets of the 10 largest U.S. commercial passenger carriers is 
years and standard deviation is 
years.
Sample size : 
Let X be the random variable that represents the age of fleets.
We assume that the ages of the fleets of the 10 largest U.S. commercial passenger carriers are normally distributed.
For z-score,
For x=14
By using the standard normal distribution table , the probability that the average age of these 40 airplanes is at least 14 years old will be :-
Hence, the required probability = 0.0129
Answer:
$12.79 + $6.75 = $19.54
Step-by-step explanation:
4 prints
12x = 2.5x + 38
at 4, these expressions are equal
Answer:
Step-by-step explanation:
Hello!
X: the lifespan of a new computer monitor of Glotech.
The average life is μ= 85 months and the variance δ²= 64
And a sample of 122 monitors was taken.
You need to calculate the probability that the sample mean is greater than 86.6 months.
Assuming that the variable has a normal distribution X~N(μ;δ²), then the distribution of the sample mean is X[bar]~N(μ;δ²/n)
To calculate this probability you have to work using the sampling distribution and the following formula Z= (X[bar]-μ)/δ/√n ~N(0;1)
P(X[bar]>86.6)= 1 - P(X[bar]≤86.6)
1 - P(Z≤(86.6-85)/(8/√122))= 1 - P(Z≤2.21)= 1 - 0.98645= 0.013355
The probability of the sample mean is greater than 0.013355
I hope this helps!
