Answer: the probability is 0.364
Step-by-step explanation:
Since the compressive strength of samples of cement can be modeled by a normal distribution, we would apply the formula for normal distribution which is expressed as
z = (x - µ)/σ
Where
x = the compressive strength of samples of cement.
µ = mean compressive strength.
σ = standard deviation
From the information given,
µ = 6000 kg/cm²
Variance = 8100
σ = √variance = √8100 = 90
We want to find the probability that the compressive strength is between 5900 and 6000 kg/cm2. It is expressed as
P(5900 ≤ x ≤ 6000)
For x = 5900,
z = (5900 - 6000)/90 = - 1.11
Looking at the normal distribution table, the probability corresponding to the z score is 0.136
For x = 6000,
z = (6000 - 6000)/90 = 0
Looking at the normal distribution table, the probability corresponding to the z score is 0.5
P(5900 ≤ x ≤ 6000) = 0.5 - 0.136
P(5900 ≤ x ≤ 6000) = 0.364
