The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 165 and a variance of 16. The material is considered defective if the breaking strength is less than 156.2 pounds. What is the probability that a single, randomly selected piece of material will be defective? (You may need to use the standard normal distribution table. Round your answer to four decimal places.)
My teacher assigned this problem online. I've tried finding this answer by plugging the numbers into the formula and using the standard normal distribution table, but I'm not coming up with the right number. This is the last problem on my last homework assignment before the semester ends. Please help me!
mean = 165
defective if <156.2
variance = 16
so sigma = sqrt 16 = 4 (did you notice that?)
then use Davidlane
mean = 165
defective if <156.2
variance = 16
so sigma = sqrt 16 = 4 (did you notice that?)
then use Davidlane
http://davidmlane.com/hyperstat/z_table.html
I get 0.0139
or 1.39 percent
To find the probability that a single, randomly selected piece of material will be defective, we need to find the area under the normal distribution curve to the left of the cutoff value of 156.2 pounds.
To start, we need to convert the given values to the standard normal distribution (also known as the Z-distribution), which has a mean of 0 and a standard deviation of 1.
To convert the breaking strength of 156.2 pounds to the standard normal distribution, we use the formula:
Z = (X - μ) / σ
where Z is the standard score, X is the value we want to convert, μ is the mean, and σ is the standard deviation.
In this case, X = 156.2, μ = 165, and σ = sqrt(16) = 4.
Using these values, we can calculate the Z score:
Z = (156.2 - 165) / 4 = -2.2
Next, we need to find the probability corresponding to this Z score. We can use the standard normal distribution table to find this probability.
The table provides the area to the left of a given Z score. To find the probability to the left of -2.2, we search for the closest value in the table. However, since the table typically provides values rounded to two decimal places, we might not find an exact match for -2.2. In such cases, we need to use the closest value available.
From the standard normal distribution table, we find that the closest value is -2.25. The corresponding cumulative probability is 0.0122.
Since the standard normal distribution is symmetric, the probability of a Z score less than -2.2 is the same as the probability of a Z score greater than 2.2. Therefore, we can use the symmetry property and subtract the cumulative probability of 2.2 (which is also 0.0122) from 1:
P(Z < -2.2) = 1 - P(Z > 2.2) = 1 - 0.0122 = 0.9878
So, the probability that a single, randomly selected piece of material will be defective is approximately 0.9878, or 98.78% (rounded to four decimal places).