In a right triangle, the hypotenuse is 15 cm and a leg is 11 cm. In a similar right triangle, the
hypotenuse is 9 cm. Find the lengths of the legs of
the smaller triangle. Write your answers to the
nearest tenth.

a) Probability of at most 20 mm = P(x ≤ 20) = 0.0526

Probability of less than 20 mm = P(x < 20) ≈ 0.0526

b) The 75th percentile of the defect length distribution = 36.988 mm

c) The 15th percentile of the defect length distribution = 24.334 mm

d) The values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%

= 22.513 mm and 41.487 mm respectively.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 32 mm

Standard deviation = σ = 7.4 mm

(a) The probability that defect length is at most 20 mm and Less than 20 mm

The required probabilities are P(x ≤ 20) and P(x < 20)

To find these, we first standardize/normalize 20 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (20 - 32)/7.4 = - 1.62

The required probabilities

We'll use data from the normal probability table for these probabilities

P(x ≤ 20) = P(z ≤ -1.62) = 0.05262

P(x < 20) = P(z < -1.62) ≈ P(z ≤ -1.62) = 0.05262

(b) The 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths from the largest 25%

Let the defect length be x'

P(x ≤ x') = 0.75

The z-score of this defect length = z'

P(z ≤ z') = 0.75

Using the normal distribution tables,

z' = 0.674

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

0.674 = (x' - 32)/7.4

x' = 36.9876 mm = 36.988 mm to 3 d.p

(c) The 15th percentile of the defect length distribution

Let the defect length be x'

P(x ≤ x') = 0.15

The z-score of this defect length = z'

P(z ≤ z') = 0.15

Using the normal distribution tables,

z' = -1.036

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

-1.036 = (x' - 32)/7.4

x' = 24.3336 mm = 24.334 mm to 3 d.p

(d) What values separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%? (Round your answers to three decimal places.) smallest 10% mm largest 10% mm

Let the two of them be x' and x" and their z-scores be z' and z"

P(x' ≤ x ≤ x") = 0.8

But it is explained that the 0.8 is the middle one, with 10% of the smallest values in the lowest range of defect lengths and another 10% of topmost values in the upper range of defect length l

P(x ≤ x') = 0.10

P(x ≤ x") = 0.10 + 0.80 = 0.90

Taking it one at a time

P(x ≤ x') = 0.10

The z-score of this defect length = z'

P(z ≤ z') = 0.10

Using the normal distribution tables,

z' = - 1.282

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

-1.282 = (x' - 32)/7.4

x' = 22.5132 mm = 22.513 mm to 3 d.p

P(x ≤ x") = 0.90

The z-score of this defect length = z"

P(z ≤ z") = 0.90

Using the normal distribution tables,

z' = 1.282

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

1.282 = (x' - 32)/7.4

x' = 41.4868 mm = 41.487 mm to 3 d.p

Hope this Helps!!!