8.value:
10.00 points
The fracture strength of a certain type of manufactured glass is normally distributed with a mean of 509 MPa with a standard deviation of 17 MPa.
(a)
What is the probability that a randomly chosen sample of glass will break at less than 509 MPa? (Round your answer to 2 decimal places.)
In a normally distributed random variable, the probability of X≤μ is exactly 0.5, since the normal distribution curve is symmetrical.
To find the probability that a randomly chosen sample of glass will break at less than 509 MPa, we need to calculate the area under the normal distribution curve to the left of 509 MPa. This represents the probability of the fracture strength being less than 509 MPa.
To do this, we can use the standard normal distribution table or a statistical calculator that provides the cumulative distribution function (CDF) for the normal distribution.
Step-by-step solution using the standard normal distribution table:
1. Standardize the value 509 MPa using the formula:
Z = (X - μ) / σ
where X is the value, μ is the mean, and σ is the standard deviation.
In this case, X = 509 MPa, μ = 509 MPa, and σ = 17 MPa.
Z = (509 - 509) / 17 = 0
2. Look up the standardized value Z = 0 in the standard normal distribution table. The table provides the cumulative probability for values up to Z.
The cumulative probability for Z = 0 is 0.5000.
3. The probability that a randomly chosen sample of glass will break at less than 509 MPa is equal to the cumulative probability at Z = 0.
Therefore, the probability is 0.5000 or 50%.
Thus, the probability that a randomly chosen sample of glass will break at less than 509 MPa is 50%.