Calculate the probability that X lies between 85 and 105 cm for the problem outlined in Example 4.1.

In this example we are required to calculate P (85 ≤ X ≤ 105) which represents the area shaded in Figure 4.5.

The value of P (85 ≤ X ≤ 105) can be calculated using Excel’s NORM.DIST () function.

Data from Example 4.1

A manufacturing firm quality assures components manufactured and historically the length of a tube is found to be normally distributed with a population mean of 100 cm and a population standard deviation of 5 cm.

Calculate the probability that a random sample of one tube will have a length of at least 110 cm.

From the information provided we define X has the tube length in centimetres and population mean μ = 100 and standard deviation σ = 5. This can be represented using the notation X ~ N (100, 5²).

The problem we have to solve is to calculate the probability that 1 tube will have a length of at least 110 cm.

This can be written as P(X ≥ 110) and is represented by the shaded area illustrated in Figure 4.2.

This problem can be solved by using the Excel function NORM.DIST (X, μ, σ², TRUE). This function calculates the area illustrated in Figure 4.3.

Transcribed Image Text:

Normal curve f(x) PDF = NORM.DIST() = p(85 <= x <= 105) = 0.839995 X₂=85 μ= 100 X₁ = 105 X Figure 4.5