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Let X be a normal random variable with mean 12 and variance 4. Find the value of c such that P(X>c)=0.10.

This question aims to find the value of $c$ given the probability distribution of a random variable $X$.

In probability theory, a random variable is regarded as a real-valued function that is defined over a sample space of a random experiment. In other words, it describes the outcome of an experiment numerically. Random variables can be categorized as discrete and continuous. The discrete random variables are one with specified values and the continuous random variables take any value within an interval.

Let $X$ be a continuous random variable. The probability distribution of $X$ assigns the probabilities to intervals on the $x-$axis with the aid of the probability density function $f(x)$. The area of the region bounded above by the graph of the equation $y=f(x)$, below by the $x-$axis, and on the left and right by the vertical lines through $a$ and $b$ is equal to the probability that a randomly chosen value of $X$ is in the interval $(a,b)$.

Expert Answer

Let $\mu=12$ and $\sigma^2=4$ be the variance of the random variable $X$.

Since $P(X>c)=0.10$

So, $P(X>c)=1-P(X\leq c)=0.10$

or,  $P(X\leq c)=1-0.10=0.90$

Also, $P(X\leq c)=P\left(Z\leq \dfrac{x-\mu}{\sigma}\right)$

Here, $x=c,\, \mu=12$ and $\sigma=\sqrt{4}=2$

Therefore, $P\left(Z\leq \dfrac{x-\mu}{\sigma}\right)=P\left(Z\leq \dfrac{c-12}{2}\right)=0.90$

$\Phi\left(\dfrac{c-12}{2}\right)=0.90$

So, by inverse use of $z-$ table, when $\Phi(z)=0.90$ then $z\approx 1.28$. And hence:

$\dfrac{c-12}{2}=1.28$

$c-12=2.56$

$c=14.56$

Example 1

Assume $X$ as a normally distributed random variable with variance $\sigma^2=625$ and mean $\mu=9$. Determine $P(65<X<70)$ and $P(78<X<80)$.

Solution

Here, $\mu=9$ and $\sigma=\sqrt{625}=25$

Therefore, $P(65<X<70)=P\left(\dfrac{x-\mu}{\sigma}<Z<\dfrac{x-\mu}{\sigma}\right)$

$P\left(\dfrac{65-9}{25}<Z<\dfrac{70-9}{25}\right)=P(2.24<Z<2.44)$

$P(2.24<Z<2.44)=P(0<Z<2.44)-P(2.24<Z<0)=0.9927-0.9875=0.00520$

And,  $P(78<X<80)=P\left(\dfrac{x-\mu}{\sigma}<Z<\dfrac{x-\mu}{\sigma}\right)$

$P\left(\dfrac{78-9}{25}<Z<\dfrac{80-9}{25}\right)=P(2.76<Z<2.84)$

$P(2.76<Z<2.84)=P(0<Z<2.84)-P(2.76<Z<0)=0.9977-0.9971=0.00060$

Example 2

A radar unit is used to monitor the speed of vehicles on a highway. The average speed is $105\, km/hr$, with a standard deviation of $5\, km/hr$. What is the likelihood that a vehicle chosen at random is traveling faster than $109\, km/hr$?

Solution

Here, $\mu=105$ and $\sigma=5$

To find:  $P(X>109)$

Now, $P(X>109)=P\left(Z>\dfrac{109-105}{5}\right)$

$P(Z>0.8)=1-P(Z\leq 0.8)=1-0.7881=0.2119$

Geogebra export

Area under the normal curve for $P(X\geq 109)$

Example 3

A large number of students took a Mathematics test. The mean and standard deviation of the final grades are $60$ and $12$, respectively. Assume the grades to be normally distributed, what percent of students scored more than $70$?

Solution

Formulate the problem as:

$P(X>70)=P\left(Z>\dfrac{x-\mu}{\sigma}\right)$

Here, $x=70,\, \mu=60$ and $\sigma=12$.

Therefore, $P\left(Z>\dfrac{x-\mu}{\sigma}\right)=P\left(Z>\dfrac{70-60}{12}\right)=P(Z>0.83)$

$P(Z>0.83)=1-P(Z\leq 0.83)=1-0.7967=0.2033$

The percentage of students who scored more than $70$ is $20.33\%$.

Images/mathematical drawings are created with GeoGebra.

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