Respuesta :
Answer:
the approximate probability that the total weight of the passengers exceeds 4668 pounds is 0.0089
Step-by-step explanation:
Using the central limit Theorem
z = (x - u )/Q÷ sqrt n
Where z is the z-score
x is the average weight = 4668/22 = 212.2
u is the passengers average = 194 pounds
Q is standard deviation = 36 pounds
Therefore
z= (212.2 - 194) / 36 ÷ sqrt22
= 18.2/ 36÷ 4.69
= 18.2/7.68
z=2.3698 aprx 2.37
Find the value of 2.37 z-score on the z-table to determine the approximate probability
P(z>2.37) = 0.0089 (from the z-table)
Answer:
Probability = 0.0089
Step-by-step explanation:
From the question ;
Mean (μ) = 194 pounds
Standard deviation(σ) = 36 pounds
Minimum total weight = 4688
Passengers carried = 22
Thus average weight = 4668/22 = 212.18
The z-value is given by;
Z = (x - μ) / ((σ)/√n)
Z= (212.18 - 194)/(36/√22)
Z = 18.18/7.675 = 2.37
Using the table I attached ;
P(z>2.37) = 0.0089
