Answer to Question #117253 in Statistics and Probability for Michael

Question

A machine produces bottle caps. The machine is calibrated to produce bottle caps with a mean diameter of 2.8 cm and standard deviation of 0.03 cm. Let x represent the random variable for diameter of bottle caps. Leave your answers as a percent without rounding.

Q1. What percent of bottle caps are between 2.77cm and 2.83 cm?
Q2. Find P(2.74 ≤ x ≤ 2.86 ).
Q3. What percent of bottle caps are less than 2.71 cm?
Q4. Find P ( X ≥ 2.86 )
Q5. What percent of bottle caps are exactly 2.8 cm?
Q6. If all caps that are smaller than 2.71 cm and greater than 2.89 cm are rejected / thrown out, what percent would be lost ?

Answer 1

1) P(2.77<x<2.83)=P(x<2.83)-P(x<2.77)

P(x<2.83), z= (x-μ)/σ=1 , P(x<2.83)=0.84134

P(x<2.77), z= (x-μ)/σ=-1 , P(x<2.77)=0.15866

P(2.77<x<2.83)=0.84134-0.15866=0.68268=68.27%

2) P(2.74<x<2.86)=P(x<2.86)-P(x<2.74)

P(x<2.86), z= (x-μ)/σ=2 , P(x<2.86)=0.97725

P(x<2.74), z= (x-μ)/σ=-2 , P(x<2.74)=0.02275

P(2.74<x<2.86)=0.97725-0.02275=0.9545

3) P(x<2.71), z= (x-μ)/σ=-3 , P(x<2.71)=0.00135=0.13%

4)  P ( X ≥ 2.86 ) =1-P(x<2.86)=1-0.97725=0.02275

5) P(x=2.8)=P(x<2.81)-P(x<2.79)

P(x<2.81), z= (x-μ)/σ=0.33 , P(x<2.81)=0.6293

P(x<2.79), z= (x-μ)/σ=-0.33 , P(x<2.79)=0.3707

P(x=2.8)=0.6293-0.3707=0.2586

6) P=P(x<2.71)+P(x>2.89)=1+P(x<2.71)-P(x<2.89)

P(x<2.71)=0.00135

P(x<2.89), z= (x-μ)/σ=3, P(x<2.89)=0.99865

P=1+0.00135-0.99865=0.0027=0.27%