The problem asks us to find the probability that the absolute deviation of a normally distributed random variable $X$ from its expected value is less than 0.8. We are given that the expected value (mean) is 7 and the standard deviation is 2.5. We need to write the probability density function of $X$ and calculate $P(|X - 7| < 0.8)$.

1. Problem Description

The problem asks us to find the probability that the absolute deviation of a normally distributed random variable $X$ from its expected value is less than 0.

8. We are given that the expected value (mean) is 7 and the standard deviation is 2.

5. We need to write the probability density function of $X$ and calculate $P(|X - 7| < 0.8)$.

2. Solution Steps

First, let's write the probability density function (PDF) for a normal distribution.

The PDF of a normal distribution is given by:

$f(x) = \frac{1}{\sigma \sqrt{2\pi}} e^{-\frac{(x - \mu)^2}{2\sigma^2}}$

where $\mu$ is the mean and $\sigma$ is the standard deviation. In this case, $\mu = 7$ and $\sigma = 2.5$. So the PDF is:

$f(x) = \frac{1}{2.5 \sqrt{2\pi}} e^{-\frac{(x - 7)^2}{2(2.5)^2}} = \frac{1}{2.5 \sqrt{2\pi}} e^{-\frac{(x - 7)^2}{12.5}}$

Next, we want to find $P(|X - 7| < 0.8)$. This is equivalent to finding $P(7 - 0.8 < X < 7 + 0.8)$, or $P(6.2 < X < 7.8)$.

To find this probability, we need to integrate the PDF from 6.2 to 7.8:

$P(6.2 < X < 7.8) = \int_{6.2}^{7.8} \frac{1}{2.5 \sqrt{2\pi}} e^{-\frac{(x - 7)^2}{12.5}} dx$

We can standardize the normal distribution using the z-score formula:

$z = \frac{x - \mu}{\sigma}$

For $x = 6.2$: $z_1 = \frac{6.2 - 7}{2.5} = \frac{-0.8}{2.5} = -0.32$

For $x = 7.8$: $z_2 = \frac{7.8 - 7}{2.5} = \frac{0.8}{2.5} = 0.32$

So, we need to find $P(-0.32 < Z < 0.32)$ where $Z$ is a standard normal random variable.

$P(-0.32 < Z < 0.32) = P(Z < 0.32) - P(Z < -0.32)$. Because the standard normal distribution is symmetric around 0, $P(Z < -0.32) = 1 - P(Z < 0.32)$ or $P(Z < -0.32) = P(Z > 0.32)$. Another property is $P(Z < -0.32) = P(Z > 0.32) = 1-P(Z < 0.32)$. Using a standard normal table or calculator, $P(Z < 0.32) \approx 0.6255$. Then, $P(-0.32 < Z < 0.32) = 0.6255 - (1-0.6255) = 0.6255 - 0.3745 = 0.2510$. Alternatively, $P(-0.32 < Z < 0.32) = P(Z < 0.32) - P(Z < -0.32)= P(Z < 0.32) - (1-P(Z<0.32)) = 2P(Z<0.32)-1 = 2(0.6255)-1=1.251-1 = 0.2510$.

3. Final Answer

The probability that the absolute deviation of the random variable from its expected value is less than 0.8 is approximately 0.

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0. Final Answer: 0.2510