Question

A random variable x follows a normal distribution with mean d and standard deviation o=2. It is known that x is less than 5 about 69.85% of the time. What is the mean d of this distribution?

Answer 1

Answer:

The mean of this distribution is approximately 3.96.

Step-by-step explanation:

Here's how to solve this problem using a normal distribution table.

Let $z$ be the

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

In this question, $x = 5$ and $\sigma = 2$. The equation becomes

$\displaystyle z = \frac{5 - \mu}{2}$.

To solve for $\mu$, the mean of this distribution, the only thing that needs to be found is the value of $z$. Since

The problem stated that $P(X \le 5) = 69.85\% = 0.6985$. Hence, $P(Z \le z) = 0.6985$.

The problem is that the normal distribution tables list only the value of $P(0 \le Z \le z)$ for $z \ge 0$. To estimate  $z$ from $P(Z \le z) = 0.6985$, it would be necessary to find the appropriate

Since $P(Z \le z) = 0.6985$ and is greater than $P(Z \le 0) = 0.50$, $z > 0$. As a result, $P(Z \le z)$ can be written as the sum of $P(Z < 0)$ and $P(0 \le Z \le z)$. Besides, $P(Z < 0) = P(Z \le 0) = 0.50$. As a result:

$\begin{aligned}&P(Z \le z)\\ &= P(Z < 0) + P(0 \le Z \le z) \\ &= 0.50 + P(0 \le Z \le z)\end{aligned}$.

Therefore:

$\begin{aligned}&P(0 \le Z \le z) \\ &= P(Z \le z) - 0.50 \\&= 0.6985 - 0.50 \\&=0.1985 \end{aligned}$.

Lookup $0.1985$ on a normal distribution table. The corresponding $z$-score is $0.52$. (In other words, $P(0 \le Z \le 0.52) = 0.1985$.)

Given that

• $z = 0.52$,
• $x =5$, and
• $\sigma = 2$,

Solve the equation $\displaystyle z = \frac{x - \mu}{\sigma}$ for the mean, $\mu$:

$\displaystyle 0.52 = \frac{5 - \mu}{2}$.

$\mu = 5 - 2 \times 0.52 = 3.96$.