The breaking strength (in pounds) of a certain new synthetic is normally distributed, with a mean of 165 and a variance of 9. The material is considered defective if the breaking strength is less than 157.8 pounds. What is the probability that a single, randomly selected piece of material will be defective
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21.19 %
To find the probability that a single randomly selected piece of material will be defective, we need to calculate the cumulative probability of the breaking strength being less than 157.8 pounds.
Since we know that the breaking strength follows a normal distribution with a mean of 165 and a variance of 9, we can use the Z-score formula to standardize the value of 157.8 pounds.
The Z-score formula is: Z = (X - μ) / σ
Where:
- Z is the standard score or Z-score
- X is the value we want to standardize
- μ is the mean of the distribution
- σ is the standard deviation of the distribution
In this case, X = 157.8, μ = 165, and σ = √9 = 3.
Calculating the Z-score:
Z = (157.8 - 165) / 3
Z = -7.2 / 3
Z = -2.4
Now, we can use the Z-score to find the cumulative probability using a standard normal distribution table or calculator. The cumulative probability represents the area under the curve to the left of the Z-score.
Looking up the Z-score of -2.4 in the standard normal distribution table, we find that the cumulative 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%.