The compressive strength of samples of cement can be modeled by a normal distribution with a mean of 6000 kg/cm2 and a variance of 8100 (kg/cm2 ) 2 . What is the probability that the compressive strength is between 5900 and 6000 kg/cm2?

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

its c or d   take a lucky guess my dude

40 minutes

step-by-step explanation:

add the time and multiply by 2