Question

Artificial silk fibers made from spider silk proteins have high tensile strengths. Tensile strength can be modeled by a normal distribution with mean 150 MJ/m3 and a standard deviation of 10 MJ/m3. Find the probability that a fiber will have tensile strength between 144.5 and 155.5 MJ/m3

Answer 1

Answer:

P(144.5≤X≤155.5)= 0.4176

Step-by-step explanation:

Hello!

Your study variable is:

X: Tensile strength of artificial silk fibers. (MJ/m³)

X~N(μ;σ²)

μ= 150 MJ/m³

σ= 10 MJ/m³

Since the only tabulated normal distribution is the standard normal, to obtain the probability that the tensile strength will be between 144.5 and 155.5 MJ/m³ you need to transform the study variable in terms of the standard normal distribution.

To do this transformation, known as standardization, you have to subtract the variable its mean and divide it by its standard deviation.

The first step is to present the probability symbolically:

P(144.5≤X≤155.5)= P(X≤155.5) - P(X≤144.5)

Now you standardize each term using the standard normal Z=(X-μ)/δ

P(Z≤(155.5-150)/10) - P(X≤(144.5-150)/10)

P(Z≤0.55) - P(Z≤-0.55)

The next step is to look for the values in the table to reach the cumulative probability for each of them.

The standard normal table has two sides, the left one shows the accumulated probabilities for negatives values of Z and the right one shows values of accumulated probabilities for positive values of Z. The first column shows the values of the integer plus first decimal of the Z value, the first row shows the second decimal of the Z value, by crossing them both you reach the probability value. (see attachment)

P(Z≤0.55) - P(Z≤-0.55)= 0.7088 - 0.2912= 0.4176

I hope it helps!