Question

the average of operating error due to the wear of the piston rings of a car shows a normal distribution with a mean of 30 mm and a variance of 36mm; a) Find the possibility of making a maximum of 34mm errors. b) Find the probability that operating errors are between 28-32 mm.

Answer 1

Given that, mean (μ) = 30 mm and

Variance $(\sigma^2)$ = 36 mm

=> Standard deviation $(\sigma)$ = √(36) = 6 mm

Let X ~ N(30, 6)

a) We want to find possibility of making a maximum of 34 mm errors.

That means we want to find, the probabilitiy that making at least 34 mm errors. (i.e. P(X ≥ 34)

$\\ P(X \geq 34) \\\\ = P(\frac {X -\mu}{\sigma} \geq \frac {34-30}{6}) \\\\ = P(Z \geq 0.67) \\\\ = 1 - P(Z \leq 0.67) \\\\ = 1 - 0.7486 \\\\ = 0.2514$

Therefore, required possibility is 0.2514 (OR 25.14%).

b) We want to find, P(28 < X < 32)

$\\ P(28 < X < 32) \\\\ = P(\frac {28-30}{6} < \frac {X -\mu}{\sigma} <\frac {32-30}{6}) \\\\ = P(-0.33 <Z < 0.33) \\\\ = 2*P(Z \leq 0.33) - 1 \\\\ = (2 * 0.6293) - 1 \\\\ = 1.2586 - 1 \\\\ = 0.2586$

Therefore, required probability is 0.2586