Aspire Faculty ID #11949 · Topic: NIMCET 2025 · Just now
NIMCET 2025

The scores of students in a national level examination are normally distributed with a mean of 500 and a standard deviation of 100. If the value of the cumulative distribution of the standard normal random variable at 0.5 is 0.691, then the probability that a randomly selected student scored between 450 and 500 is

Solution

Given: 
Scores are normally distributed with mean $ \mu = 500 $, 
standard deviation $ \sigma = 100 $ 
We want the probability that a student scored between 450 and 500: 
$P(450 < X < 500)$ 

Convert to $Z$-scores: 
For $X = 450$: $Z = \dfrac{450 - 500}{100} = -0.5$ 

For $X = 500$: $Z = \dfrac{500 - 500}{100} = 0$ 

Thus: 
$P(450 < X < 500) = P(-0.5 < Z < 0)$ 

Given: $P(Z < 0.5) = 0.691$ 

Using symmetry of the standard normal curve:
 $P(-0.5 < Z < 0.5) = 2(0.691 - 0.5) = 2(0.191) = 0.382$ 
So: $P(-0.5 < Z < 0) = \dfrac{0.382}{2} = 0.191$ 

Final answer: 0.191
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