Complete question is;
A safety light is designed so that the times between flashes are normally distributed with a mean of 2.50s and a standard deviation of 0.50s.
A. Find the probability that an individual time is greater than 3.00 s.
B. Find the probability that the mean for 70 randomly selected times is greater than 3.00 s.
C. Given that the light is intended to help people see an obstruction, which result is more relevant for assessing the safety of the light?
Answer:
A) 0.15866
B) 0.00000001
C) Result in part B is more important
Step-by-step explanation:
A) We are given;
Mean; μ = 2.5
Standard deviation; σ = 0.5
To find the probability that an individual time is greater than 3 seconds, we will find the z-score and then find the p-value.
z = (x¯ - μ)/σ
z = (3 - 2.5)/0.5
z = 1
From z-distribution table attached, at z = 1, p(z < 1) = 0.84134
But we are looking for P(z > 1), thus;
P(z > 1) = 1 - 0.84134
P(z > 1) = 0.15866
B) now 70 are randomly selected. Thus, the z-score will be gotten from;
z = (x¯ - μ)/(σ/√n)
z = (3 - 2.5)/(0.5/√70)
z ≈ 8.37
From online calculations, we have;
P(z < 8.37) = 0.99999999
But we need P(z > 8.37).
This is:
P(z > 8.37) = 1 - P(z < 8.37)
P(z > 8.37) = 1 - 0.99999999
P(z > 8.37) = 0.00000001
C) The result in part B is more relevant because it is dealing with a sample size of 70 as against individual one in result A.
