The random variable x is normally distributed with mean 5 and standard deviation 25.

While a high standard deviation suggests that the values are dispersed throughout a wider range, a low standard deviation suggests that the values tend to be close to the established mean.

So, using the concepts of expected value and variance of random variables, this issue can be resolved.

By the expected value principle:

E(Y) = E(2+4X) = E(2) + E(4X) = 2 + 4E(X) = 2 + 4(5) = 22

[tex]\mathrm{E}(\mathrm{Y})=\mathrm{E}(2+4 \mathrm{X})=\mathrm{E}(2)+\mathrm{E}(4 \mathrm{X})=2+4 \mathrm{E}(\mathrm{X})=2+4(5)=22[/tex]

Then, μy = 22

In terms of variance concept:
Var(Y) = Var(2 + 4x) = Var(2) + Var(4x) + 2Cov(2, 4x)

Cov(2, 4x) = 0

Var(2) = 0

Hence:
Var(Y) = Var(4x) = 4²Var(X) = 16σ²ₓ = 16(25²) = 10000

Standard deviation(Y) = √10000 = 100

Therefore, (C) the standard deviation is 100 and the mean is 22.

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Complete question:
The random variable X is normally distributed with a mean of 5 and a standard deviation of 25. The random variable Y is defined by Y = 2 + 4X. What are the mean and the standard deviation of Y? The mean is 20 and the standard deviation is 102.

A) The mean is 20 and the standard deviation is 50.

B) The mean is 22 and the standard deviation is 102.

C) The mean is 22 and the standard deviation is 100.

D) The mean is 22 and the standard deviation is 50.