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Assume the lengths of cracks in the outer wall of the Swan Building are normally

ID: 3316345 • Letter: A

Question

Assume the lengths of cracks in the outer wall of the Swan Building are normally distributed, with a mean length of 28 inches and a standard deviation of 2 inches. A crack is selected at random, to answer the following: 2. a- Probability of the crack measuring less than 26 inches? b- Probability of the crack measuring between 25 and 31 inches? c- Probability of the crack measuring more than 30 inches? d- Probability of the crack measuring exactly 29 inches? e- How long do you expect the longest 11% of cracks on the building would measure?

Explanation / Answer

Length of carcks are normally distributed with mean = 28 and standard deviation = 2

Let X be the length of cracks.

X ~ N(28, 4)

(a)

Probability of carck measuring less than 26

= P(X < 26)

=  pnorm(26, mean = 28, sd = 2)

= 0.1586553

(b)

Probability of crack measuring between 25 and 31 inches

= P(25 < X < 31)

= P(X < 31) - P(X < 25)

= pnorm(31, mean = 28, sd = 2) - pnorm(25, mean = 28, sd = 2)

= 0.8663856

(c)

Probability of crack measuring more than 30

= P(X > 30)

= 1 - P(X <= 30)

= 1 - pnorm(30, mean = 28, sd = 2)

= 0.1586553

(d)

Probability of crack measuring exactly 29 inches

= P(X = 29)

= 0

since probability of crack measuring exactly equal to some fixed length = 0

(e)

The length of the longest 11% cracks would be equal to y

such that

P(X > y) = 0.11

=> P(X <= y) = 0.89

=> y = qnorm(0.89, mean = 28, sd = 2)

=  30.45306 inches

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