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Bottle Caps 

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.  

What percent of bottle caps are between 2.77 cm and 2.83 cm?  

A:  

Find P(2.74 ≤ x ≤ 2.86) 

A: 

What percent of bottle caps are less than 2.71 cm?  

A:  

Find P(X ≥ 2.86) 

A:  

What percent of bottle caps are exactly 2.8 cm? 

A: 

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? 

A: 

 

Men's Shoe Designer 

A men's shoe designer is willing to accommodate 95% of the male population.  If the mean foot length is 25.25 cm and the standard deviation is 3.875 cm, answer the following: 

What is the smallest foot length the designed is willing to accommodate? 

A:  

What is the largest foot length the designed is willing to accommodate? 

A: 

1 year ago

Answered By Sara S

$\mu$ $$\mu$$


1 year ago

Answered By Sara S

Hi there, 

 

Bottle Caps :

We use z-score method. 

Let  $\mu$μ  = mean = 2.8 and  $\sigma$σ = standard deviation = 0.03.

Then, we have z =  $\frac{X-\mu}{\sigma}$Xμσ . Accordingly, we have 

P((2.74- $\mu$μ )/ $\sigma$σ  $\le$  z  $\le$ (2.86- $\mu$μ )/ $\sigma$σ ) = P((2.74 - 2.8)/0.03  $\le$$\le$ (2.86 - 2.8)/0.03)=

P(-0.06/0.03   $\le$   z  $\le$ 0.06/0.03)= P(-2  $\le$ z  $\le$ 2) = P(z  $\le$ 2) - P(z $\le$ -2). 

Form the z-table, we have 

P(z $\le$ 2 )= 0.9772 and P(z $\le$ -2) = 0.0228. 

Therefore, we have 

P(2.74  $\le$  X  $\le$ 2.86) = 0.9772 - 0.0228 = 0.9544

 

For all other parts, use the same method and let me know if you have any questions. 


1 year ago

Answered By Sara S

Correction: 

In the second line, z =  $\frac{X-\mu}{\sigma}$Xμσ  


1 year ago

Answered By Sara S

Men's Shoe Designer:

Based on the 95% Rules, we have:

Movin 2 standard deviations to right of the mean and also to the left of the mean, we cover 95% of values. In the other word, 

P( $\mu-2\sigma\le$μ$\le\mu+2\sigma$μ+ )  $\approx$ 95%

Since the guy is focusing on 95% of the Male population, we conclude:

 the smallest foot length the designed is willing to accommodate = $\mu-2\sigma$μ  = 25.25 - 2(3.875) = 17.5

 

and

 

 the largest foot length the designed is willing to accommodate  $\approx$  $\mu+2\sigma$μ+  = 25.25+2(3.875) = 33

 

Hope this helps!

Good Day!