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Horn lengths of Texas longhorn cattle are normally distributed.  The mean horn spread is 60 inches with a standard deviation of 4.5 inches.

Calculate the range of horn lengths for the middle 95% of Texas Longhorn cattle. Show your work or explain how you got your answer.

 Dec 23, 2018
 #1
avatar+679 
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We can use the variables μ (mu) to represent the length of the horn, and σ (sigma) for the standard deviation.

 

The average horn spread:  μ = 60 inches.

The standard deviation:  σ = 4.5 inches.

 

According to the normal distribution, the range for the middle 95% of the values should be: [ μ  - 2σ,  μ + 2σ ]

Replacing the given values, we have [ 60 inches  - 2(4.5 inches), 60 inches + 2(4.5 inches) ]. 

Distributing out the values, we have [ 60 inches  - 9 inches, 60 inches + 9 inches ].


Our solution is [\(\boxed{ 51 \text{inches,} 69 \text{inches} }\)]. (with the [] brackets)

 

- PM

 Dec 23, 2018
edited by PartialMathematician  Dec 23, 2018
 #2
avatar+98130 
+1

Nice, PM....!!!!

 

BTW.....we don't need the triangle, at all....

 

 

cool cool cool

CPhill  Dec 23, 2018
 #3
avatar+679 
0

Oh yeah, I just realized that I never used it. Let me remove it. Thanks, CPhill! laugh

PartialMathematician  Dec 23, 2018

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