The diameter of a pipe is normally distributed, with a mean of 0.4 inch and a variance of 0.0009. What is the probability that the diameter of a randomly selected pipe will exceed 0.418 inch

P(X > 0.418) = 0.27425

Step-by-step explanation:

We are given;

Population mean; μ = 0.4 inch

Variance; Var = 0.0009

Sample mean; x¯ = 0.418

First of all let's get the standard deviation which is;

σ = √Var

σ = √0.0009

σ = 0.03

Now, z - score formula is;

z = (x¯ - μ)/σ

z = (0.418 - 0.4)/0.03

z = 0.6

We want to find the probability that the diameter of a randomly selected pipe will exceed 0.418 inch. Thus;

P(X > 0.418) = 1 - P(X ≤ 0.418)

From z-distribution table attached, P(X ≤ 0.418) = P(Z) = 0.72575

Thus;

P(X > 0.418) = 1 - 0.72575

P(X > 0.418) = 0.27425

Answer: d. 39 cmin the diagram, the approximate length of the minor arc de is 39 centimetres.step-by-step explanation: the formula for the length of an arc in a circle is as displayed below.let the length of the minor arc de = xx = 2( pi )r × angle of sector / 360°from this, we will substitute the values given from the problem into the formula to obtain the length of the minor arc de.x = 2( pi )r × angle of sector / 360°r = radius = 25cmpi as an approximate value = 3.14angle of sector = 120°x = 2( 3.14 )( 25 ) × 120° / 360°x = 157 × 1 / 4x = 39.25x = 39 ( rounded to the nearest whole number as an approximate value )