Answer:
About 84.13%
Step-by-step explanation:
Using gaussian distribution function
f(t)=[tex]\frac{1}{\sqrt{2\pi\sigma^2 } }e-(t-t_a_v_g)^2/2\sigma^2[/tex]
T- 60
sigma- 10
we then consider the range between [0,70)
=.841345 or 84.13%
Answer:
Step-by-step explanation:
Since the time required to finish the test is normally distributed, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = time taken to finish test.
µ = mean time
σ = standard deviation
From the information given,
µ = 60 minutes
σ = 10 minutes
We want to find the probability that a student will finish the test in less than 70 minutes. It is expressed as
P(x ≤ 70)
For x = 70,
z = (70 - 60)/10 = 1
Looking at the normal distribution table, the probability corresponding to the z score is 0.8413
Therefore,
P(x ≤ 70) = 0.8413
