A company that produces rivets used by commercial aircraft manufactures knows that the shearing strength (force required to breaking) of its rivets is a major concern. The company believes the shearing strength of its rivets is normally distributed with a mean of 925 lbs and a standard deviation of 18 lbs.
A. If the company is correct, what percentage of its rivets have a shearing strength greater than 900 lbs?
B. What is the upper bound for the shearing strength of the weakest 1% of its rivets?
C. If one rivet is randomly selected from all of the rivets, what is the probability that it will require a force of at least 920 lbs to break it?
A. Z = (score-mean)/SD
Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability of the Z score. Multiply by 100.
B. Use same table to find Z for P = .01. Insert in above equation and calculate.
C. Process same as A, but don't multiply by 100.
To solve these problems, we can use the properties of the normal distribution and the standard normal distribution (also known as the Z-distribution). By converting the given values to Z-scores, we can calculate probabilities and values based on the standard normal distribution.
A. To find the percentage of rivets with a shearing strength greater than 900 lbs, we need to calculate the probability of a Z-score greater than the Z-score corresponding to 900 lbs.
The Z-score formula is:
Z = (X - μ) / σ
where X is the observed value, μ is the mean, and σ is the standard deviation.
In this case, X = 900 lbs, μ = 925 lbs, and σ = 18 lbs.
Calculating the Z-score:
Z = (900 - 925) / 18 = -1.39
We can then find the corresponding probability using a Z-table or a statistical calculator. Looking up the Z-score of -1.39 in a Z-table or using a calculator, we find that the probability is approximately 0.0823 or 8.23%.
Therefore, approximately 8.23% of rivets have a shearing strength greater than 900 lbs.
B. To find the upper bound for the shearing strength of the weakest 1% of rivets, we need to find the value corresponding to the Z-score that encloses 1% of the distribution's area. This is often referred to as the "critical Z-score" or "Z-critical value."
Since we want to find the upper bound, we need to find the Z-score that corresponds to a probability of 1% in the upper tail of the distribution.
Using a Z-table or a statistical calculator, we can find this Z-score. The Z-score that corresponds to a probability of 1% in the upper tail is approximately 2.33.
We can then use the Z-score formula to find the shearing strength associated with this Z-score:
Z = (X - μ) / σ
Since we are looking for the upper bound, X is the unknown value, μ is the mean, σ is the standard deviation, and Z is the critical value of 2.33.
2.33 = (X - 925) / 18
Solving for X:
18 * 2.33 = X - 925
41.94 = X - 925
X = 925 + 41.94
X ≈ 966.94
Therefore, the upper bound for the shearing strength of the weakest 1% of rivets is approximately 966.94 lbs.
C. To find the probability that a randomly selected rivet will require a force of at least 920 lbs to break it, we need to calculate the probability of a Z-score greater than the Z-score corresponding to 920 lbs.
Using the Z-score formula:
Z = (X - μ) / σ
In this case, X = 920 lbs, μ = 925 lbs, and σ = 18 lbs.
Calculating the Z-score:
Z = (920 - 925) / 18 = -0.28
Using a Z-table or a statistical calculator, we can find the probability associated with a Z-score of -0.28. The probability is approximately 0.3897 or 38.97%.
Therefore, the probability that a randomly selected rivet will require a force of at least 920 lbs to break it is approximately 38.97%.