Respuesta :
Answer:
Option C) 0.25
Step-by-step explanation:
We are given the following information in the question:
Mean, μ = 22 minutes
Standard Deviation, σ = 11.9 minutes
We are given that the distribution of response time is a bell shaped distribution that is a normal distribution.
Formula:
[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]
P(response times is over 30 minutes)
P(x > 30)
[tex]P( x > 30) = P( z > \displaystyle\frac{30 - 22}{11.9}) = P(z > 0.6722)[/tex]
[tex]= 1 - P(z \leq 0.6722)[/tex]
Calculation the value from standard normal z table, we have,
[tex]P(x > 30) = 1 - 0.749= 0.251 = 25.1\%[/tex]
Thus, the correct answer is
Option C) 0.25
Answer:d is the correct option.
Step-by-step explanation:
Since its response time to fires is approximately 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 response time
s = standard deviation
From the information given,
u = 22 minutes
s = 11.9 minutes
We want to find the proportion of their response times that is over 30 minutes. It is expressed as
P(x > 30) = 1 - P(x ≤ 30)
For x = 30
z = (30 - 22)/11.9 = 0.67
Looking at the normal distribution table, the probability corresponding to the z score is 0.7486
P(x > 30) = 1 - 0.7486 = 0.25
