Answer:
Explanation:
Suppose the time required for an auto shop to do a tune-up is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - u)/s
Where
x = points scored by students
u = mean time
s = standard deviation
From the information given,
u = 102 minutes
s = 18 minutes
1) We want to find the probability that a tune-up will take more than 2hrs. It is expressed as
P(x > 120 minutes) = 1 - P(x ≤ 120)
For x = 120
z = (120 - 102)/18 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.8413
P(x > 120) = 1 - 0.8413 = 0.1587
2) We want to find the probability that a tune-up will take lesser than 66 minutes. It is expressed as
P(x < 66 minutes)
For x = 66
z = (66 - 102)/18 = - 2
Looking at the normal distribution table, the probability corresponding to the z score is 0.02275
P(x < 66 minutes) = 0.02275