The thickness of a flange on an aircraft component is uniformly distributed between 0.95 and 1.05 millimeters.
(a) Determine the cumulative distribution function of flange thickness.
(b) Determine the proportion of flanges that exceeds 1.02 millimeters.
(c) What thickness is exceeded by 90% of the flanges?
(d) Determine the mean and variance of flange thickness.
The correct answer and explanation is :
We are given that the flange thickness ((X)) is uniformly distributed between 0.95 and 1.05 mm. This is a continuous uniform distribution, denoted as:
[
X \sim U(a, b)
]
where (a = 0.95) and (b = 1.05).
(a) Cumulative Distribution Function (CDF)
The probability density function (PDF) of a uniform distribution is:
[
f(x) = \frac{1}{b – a} = \frac{1}{1.05 – 0.95} = 10, \quad \text{for } 0.95 \leq x \leq 1.05
]
The cumulative distribution function (CDF) is:
[
F(x) =
\begin{cases}
0, & x < 0.95 \ \frac{x – 0.95}{0.10}, & 0.95 \leq x \leq 1.05 \ 1, & x > 1.05
\end{cases}
]
(b) Proportion of Flanges Exceeding 1.02 mm
We need (P(X > 1.02)), which is:
[
P(X > 1.02) = 1 – F(1.02)
]
[
F(1.02) = \frac{1.02 – 0.95}{0.10} = 0.7
]
[
P(X > 1.02) = 1 – 0.7 = 0.3
]
So, 30% of flanges exceed 1.02 mm.
(c) Thickness Exceeded by 90% of Flanges
We find (x) such that:
[
P(X > x) = 0.90
]
[
F(x) = 1 – 0.90 = 0.10
]
[
\frac{x – 0.95}{0.10} = 0.10
]
[
x = 0.95 + (0.10 \times 0.10) = 0.96
]
So, 90% of flanges exceed 0.96 mm.
(d) Mean and Variance
For a uniform distribution:
[
E[X] = \frac{a + b}{2} = \frac{0.95 + 1.05}{2} = 1.00
]
[
Var(X) = \frac{(b – a)^2}{12} = \frac{(1.05 – 0.95)^2}{12} = \frac{0.01}{12} = 0.000833
]
Explanation (300 words)
The uniform distribution describes a scenario where each outcome in an interval is equally likely. Here, the flange thickness is uniformly distributed between 0.95 and 1.05 mm, meaning any thickness in this range has an equal probability of occurring.
To determine the cumulative distribution function (CDF), we integrate the probability density function (PDF). The CDF helps compute probabilities for any given range. Using it, we find that the proportion of flanges exceeding 1.02 mm is 0.30, meaning 30% of flanges have a thickness greater than 1.02 mm.
Next, we determine the thickness that is exceeded by 90% of the flanges. By solving ( F(x) = 0.10 ), we find that 90% of flanges exceed 0.96 mm. This is useful in quality control to set tolerance limits.
The mean and variance describe the central tendency and spread of the thickness distribution. The mean, ( E[X] = 1.00 ), represents the expected thickness. The variance, ( 0.000833 ), measures the spread around the mean, indicating minimal variability.
Understanding these statistical properties is crucial for ensuring consistent manufacturing quality in aerospace engineering. Components must adhere to strict tolerances, and statistical methods help assess their reliability.