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# 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%

Last updated date: 18th Sep 2024
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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 (ยต) and standard deviation, sigma (๐). The graph so obtained is symmetrical.

Given
Scores on test are normally distributed
Mean (ยต) = 500
Standard deviation (๐) = 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
$\Rightarrow Z\, = \dfrac{{x - \mu }}{\sigma }$
For x = 585
$\Rightarrow Z\, = \,\dfrac{{585\, - \,500}}{{100}} \\ \Rightarrow Z = \,\dfrac{{85}}{{100}} \\ \Rightarrow Z\, = \,0.85 \\$ 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 = $\dfrac{{0.8023}}{{100}}\, = \,80.23\,\%$
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.