The given function is a normal distribution function, which is given by:

f(t) = (1/√(2πσ^2)) * e^(-(t-μ)^2 / 2σ^2)

where μ is the mean and σ is the standard deviation.

The points of inflection of the normal curve are at μ - σ and μ + σ.

Given that the function is f(t) = e^((t-10)^2 / 32) / √(4π), we can see that μ = 10 and σ = √(32/2) = 4√2.

Therefore, the points of inflection are at 10 - 4√2 and 10 + 4√2.

These values do not match any of the given options, so the correct answer is (d) none of these.

Question

The given function is a normal distribution function, which is given by:

f(t) = (1/√(2πσ^2)) * e^(-(t-μ)^2 / 2σ^2)

where μ is the mean and σ is the standard deviation.

The points of inflection of the normal curve are at μ - σ and μ + σ.

Given that the function is f(t) = e^((t-10)^2 / 32) / √(4π), we can see that μ = 10 and σ = √(32/2) = 4√2.

Therefore, the points of inflection are at 10 - 4√2 and 10 + 4√2.

These values do not match any of the given options, so the correct answer is (d) none of these.

Knowee AI · Accepted Answer

The given function is a normal distribution function, which is given by:

f(t) = (1/√(2πσ^2)) * e^(-(t-μ)^2 / 2σ^2)

where μ is the mean and σ is the standard deviation.

The points of inflection of the normal curve are at μ - σ and μ + σ.

Given that the function is f(t) = e^((t-10)^2 / 32) / √(4π), we can see that μ = 10 and σ = √(32/2) = 4√2.

Therefore, the points of inflection are at 10 - 4√2 and 10 + 4√2.

These values do not match any of the given options, so the correct answer is (d) none of these.

The points of inflexion of the normal curve( )2- t-101 32f(t) = e4 2π are(a) 6, 14(b) 5,15(c) 4,16(d) none of these