the breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 125 and a variance of 25. The material is considered defective if the breaking strength is less than 113 pounds. What is the probability that a single, randomly selected piece of material will be defective?
P(x < 113)
mean = 125
Standard deviation = sqrt(variance)
Standard deviation = 5
z = (113-125)/5
z = -12/5
z = -2.4
P ( z < -2.4) = 0.0082
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 where the breaking strength is less than 113 pounds.
Step 1: Calculate the standard deviation (σ) from the variance (σ^2):
σ = √(variance) = √(25) = 5
Step 2: Calculate the z-score for the breaking strength of 113 pounds:
z = (X - μ) / σ
where X is the breaking strength value, μ is the mean, and σ is the standard deviation.
z = (113 - 125) / 5 = -12 / 5 = -2.4
Step 3: Use a z-table or a statistical calculator to find the probability corresponding to the z-score of -2.4. This probability represents the area to the left of the z-score on the standard normal distribution curve.
Looking up the z-score of -2.4 on a standard normal distribution table, we find that the probability is approximately 0.0082.
Therefore, the probability that a single, randomly selected piece of material will be defective is approximately 0.0082, or 0.82%.