+0

# Help 18

0
107
3
+4677

Help 18

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
+812
+2

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
+94275
+1

Nice, PM....!!!!

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

CPhill  Dec 23, 2018
#3
+812
+3

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

PartialMathematician  Dec 23, 2018