A random variable $X$ is normally distributed with parameters $a = 350$ and $\sigma = 10$. We need to: (a) Write the probability density function $f(x)$. (b) Sketch the graph of $f(x)$. (c) Find the expected value, variance, and standard deviation of $X$. (d) Find the probability that $340 \le X \le 365$.

1. Problem Description

A random variable $X$ is normally distributed with parameters $a = 350$ and $\sigma = 10$. We need to:

(a) Write the probability density function $f(x)$.

(b) Sketch the graph of $f(x)$.

(c) Find the expected value, variance, and standard deviation of $X$.

(d) Find the probability that $340 \le X \le 365$.

2. Solution Steps

(a) The probability density function (PDF) for a normal distribution is given by:

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

In this case, $\mu = a = 350$ and $\sigma = 10$. Plugging these values into the formula, we get:

$$f(x) = \frac{1}{10\sqrt{2\pi}}e^{-\frac{(x-350)^2}{2(10^2)}}$$

$$f(x) = \frac{1}{10\sqrt{2\pi}}e^{-\frac{(x-350)^2}{200}}$$

(b) The graph of $f(x)$ is a bell curve centered at $x = 350$. The peak of the curve is at $x = 350$, where $f(350) = \frac{1}{10\sqrt{2\pi}} \approx 0.03989$. The curve is symmetric around $x = 350$, and its spread is determined by the standard deviation $\sigma = 10$. The graph extends infinitely in both directions but approaches zero as $x$ moves away from

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(c) For a normal distribution with parameters $\mu = a$ and $\sigma$, the expected value $E[X]$, variance $Var(X)$, and standard deviation are:

$$E[X] = \mu = a$$

$$Var(X) = \sigma^2$$

$$SD[X] = \sigma$$

In our case, $a = 350$ and $\sigma = 10$. So,

$E[X] = 350$

$Var(X) = 10^2 = 100$

$SD[X] = 10$

(d) To find the probability $P(340 \le X \le 365)$, we need to calculate the area under the curve of $f(x)$ between $x = 340$ and $x = 365$. We can standardize $X$ by defining a new variable $Z = \frac{X - \mu}{\sigma}$, which follows a standard normal distribution with mean 0 and standard deviation

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$$P(340 \le X \le 365) = P\left(\frac{340 - 350}{10} \le Z \le \frac{365 - 350}{10}\right) = P(-1 \le Z \le 1.5)$$

$$P(-1 \le Z \le 1.5) = P(Z \le 1.5) - P(Z \le -1) = P(Z \le 1.5) - P(Z \ge 1) = P(Z \le 1.5) - [1 - P(Z \le 1)]$$

Using a standard normal distribution table (or calculator), we have:

$P(Z \le 1.5) \approx 0.9332$

$P(Z \le 1) \approx 0.8413$

$$P(-1 \le Z \le 1.5) = 0.9332 - (1 - 0.8413) = 0.9332 - 0.1587 = 0.7745$$

3. Final Answer

(a) $f(x) = \frac{1}{10\sqrt{2\pi}}e^{-\frac{(x-350)^2}{200}}$

(b) The graph is a bell curve centered at x=

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0. (c) $E[X] = 350$, $Var(X) = 100$, $SD[X] = 10$

(d) $P(340 \le X \le 365) \approx 0.7745$