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