Question

Entry to a certain University is determined by a national test .The scores on this test are normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be admitted to this university and he knows that he must score better than at least 70% of the students who took the test. Tom takes the test and scores 585. Tom does better than what percentage of students ?
A. 89.23%
B. 77.26%
C. 70.23%
D. 80.23%

Answer 1

Hint: Whenever in any situation, condition or terms are given and distribution is provided and it follows normal distribution then mean is represented as mu (µ) and standard deviation, sigma (𝛔). The graph so obtained is symmetrical.

Complete step-by-step answer:
Given
Scores on test are normally distributed
Mean (µ) = 500
Standard deviation (𝛔) = 100
For getting admission he should score better than at least 70 % of students
So he should be in the remaining 30 % of students who have scored good marks.
We will find who have scored below the marks scored by tom
 We will find probability of students who scored Marks below 585 marks
x < 585
we have to find the Z REGION
$\Rightarrow Z={\displaystyle \frac{x-\mu }{\sigma }}$
For x = 585
$$\begin{gathered} \Rightarrow Z={\displaystyle \frac{585-500}{100}} \\ \Rightarrow Z={\displaystyle \frac{85}{100}} \\ \Rightarrow Z=0.85 \end{gathered}$$ 0.85
Applying probability
 P(Z < 0.85) = P(Z < 0) + P(0< Z <0.85)
p(Z<0) = 0.5
P(0< Z <0.85) = 0.3023 { from Z table at Z values normal distribution}
P(Z < 0.85) = 0.5 + 0.3023 = 0.8023
Changing this probability into percentage = ${\displaystyle \frac{0.8023}{100}}=80.23\mathrm{\%}$
Therefore tom has to score more than 80.23%

Option (D) is correct.

Note: As it was mentioned, normal distribution so we will calculate z and probability as per condition. While calculating z values from the z table. Be extra attentive, change in a decimal value at any place can change your whole observation.