Question

Keith is a drummer who purchases his drumsticks online. When practicing with the newest pair, he notices they feel heavier thanusual. When he weighs one of the sticks, he finds that it is 1.87 oz. The manufacturer's website states that the average weight ofeach stick is 1.50 oz with a standard deviation of 0.15 oz. Assume that the weight of the drumsticks is normally distributed.What is the probability of the stick's weight being 1.87 oz or greater? Give your answer as a percentage precise to at least twodecimal places.

Answer 1

The probability of the stick's weight being 1.87 oz or greater is 0.68%

STEP - BY - STEP EXPLANATION

What to find?

Probability of the stick's weight being 1.87 oz or greater.

Given:

• Mean μ = 1.50

,

• Standard deviation σ = 0.15

To solve the problem given, we will follow the steps below:

Step 1

State the formula needed to solve the problem given.

`Z=(X-\mu)/(\sigma)`

Step 2

State the required probability.

The reqyuired probability is;

P(X ≥ 1. 87) = 1 - P(X < 1.87)

Step 3

Substitute the values into the formula and simplify.

That is;

`\begin{gathered} P\mleft(X\ge1.87\mright)=1-P\mleft(X<1.87\mright) \\ \\ =1-\text{ P(}(X-1.50)/(0.15)<(1.87-1.50)/(0.15)) \\ \\ =1-P(Z<(0.37)/(0.15)) \\ \\ =1-P(Z<2.47) \end{gathered}`

Step 4

Use the Z- distribution table to determine the value of P(Z<2.47).

Using z distribution table, looking 2.0 in the left column and 0.47 in the top most row, then selecting the intersecting cell, we get 0.9932.

Step 5

Substitute the value of P(Z<2.47) into:

P(X ≥ 1. 87) = 1 - P(Z<2.47))

= 1 - 0.9932

=0.0068

Step 6

Convert the result in step 5 to percentage by multiplying by 100.

P(X ≥ 1. 87) =0.0068 * 100 =0.68 %

Thereore, the probability of the stick's weight being 1.87 oz or greater is 0.68%