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Tamil Nadu Board of Secondary EducationHSC Commerce Class 12

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 - Business Mathematics and Statistics

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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. Raghul 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. Raghul takes the test and scores 585. Will he be admitted to this university?

Sum
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Solution

Let x denotes the scores of a national test mean

µ = 500 and standard deviation σ = 100

Standard normal variate z = `(x - mu)/sigma = (x - 5000)/100`

When x = 585

z = `(585 - 500)/100 = 85/100` = 0.85

P(X ≤ 585) = P(Z ≤ 0.85)

P(Z ≤ 0.85) = P(`-oo` < z < 0) + P(0 < z < 0.85)

= 0.5 + 0.3023

= 0.8023

For n = 100

P(Z ≤ 0.85) = 100 × 0.8023

= 80.23

∴ Raehul scores 80.23%

We can determine the scores of 70% of the students as follows:

From the table for the area 0.35

We get z1 = – 1.4(as z1 lies to left of z = 0)

Similarly z2 = 1.4

Now z1 = `(x_1 - 500)/100`

⇒  – 1.4 = `(x_1 - 500)/100`

– 1.4 × 100 = x1 – 500

⇒ x1 500 – 140

x1 = 360

Again z2 = `(x_2 - 500)/100`

⇒ – 1.4 = `(x_1 - 500)/100`

1.4 × 100 = x2 – 500

⇒ x2 = 140 + 500

= x2 = 640

Hence 70% of students score between 360 and 640

But Raghul scored 585. His score is not better than the score of 70% of the students.

∴ He will not be admitted to the university.

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Chapter 7: Probability Distributions - Miscellaneous problems [Page 171]

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Samacheer Kalvi Business Mathematics and Statistics [English] Class 12 TN Board
Chapter 7 Probability Distributions
Miscellaneous problems | Q 5 | Page 171

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