Question

The scores on standardized admissions tests are normally distributed with a mean of $500$ and a standard deviation of $100$. What is the probability that a randomly selected student will score between $400$ and $600$ on the test?
$\left( A \right)$ About $63\%$
$\left( B \right)$ About $65\%$
$\left( C \right)$ About $68\%$
$\left( D \right)$ About $70\%$

Answer 1

Hint: In this question we have been given that the scores of standardized admissions tests are normally distributed. We have the values of the mean given to us as $500$ and the standard deviation given to us as $200$, we have to find the probability that a random student has the marks in the range $400$ to $600$. We will solve this question by using the $z$-table by first finding the value of $z$ for the value of $x$, which is the desired marks which will be $400$ and $600$. We will find $z=\dfrac{\left( x-\mu \right)}{\sigma }$, where $\mu $ is the mean and $\sigma $ is the standard deviation. We will then look at the normal distribution graph and find the probability in the given range of $z$.

Complete step by step solution:
We have been given that the scores are normally distributed.
We have the mean as $500$ therefore, we can write:
$\mu =500$
And we have the standard deviation as $200$ therefore, we can write:
$\sigma =200$
Now we have to find the probability that the marks of a random student picked lies in the range $400$ to $600$.
Now the value of $z$ at $x=400$ will be:
$\Rightarrow z=\dfrac{400-500}{100}$
On simplifying, we get:
$\Rightarrow z=-1$
Now the value of $z$ at $x=600$ will be:
$\Rightarrow z=\dfrac{600-500}{100}$
On simplifying, we get:
$\Rightarrow z=1$
Now we know the property of normal distribution that the probability of $-1$ to $0$ is the same as $0$ to $1$ therefore we have to multiply by $2$ whatever is the value for $z=1$.
Now at $z=1$, we have the probability as $0.3413$
Now on multiplying by $2$, we get the probability as:
$\Rightarrow 2\times 0.3413$
On simplifying, we get:
$\Rightarrow 0.6826$, which is the required probability.
On converting into percentage, we get:
$\Rightarrow 68.26\%$, which is near to $68\%$
So, the correct answer is “Option C”.

Note: It is to be remembered that in this question we have used the normal distribution which is one common type of distribution used. There are various other probability distributions which should be remembered such as the binomial distribution, Bernoulli distribution etc. It is to be noted that the answer which we got is an approximate answer.