Question

The marks obtained by a number of students in a certain subject are approximately normally distributed with mean 65 and standard deviation 5. If 3 students are selected at random from this group, what is the probability that at least 1 of them would have scored above 75?

Answer 1

Answer on Question #44957 – Math - Statistics and Probability

The marks obtained by a number of students in a certain subject are approximately normally distributed with mean 65 and standard deviation 5. If 3 students are selected at random from this group, what is the probability that at least 1 of them would have scored above 75?

Solution

Let event $A =$ at least 1 of 3 students would have scored above 75

The complement of A is $\bar{A} =$ none of 3 students would have scored above 75

$\bar{A} = B \cap C \cap D$ is intersection of three events $B, C, D$, where

$B =$ the first student wouldn't have scored above 75 =

= "the first student would have scored less than 75";

$C =$ "the second student wouldn't have scored above 75" =

= "the second student would have scored less than 75";

$D =$ "the third student wouldn't have scored above 75" =

= "the third student would have scored less than 75".

$B, C, D$ are jointly statistically independent events.

Let $X$ represents the marks obtained by a student. $X$ is normally distributed with mean 65 and standard deviation 5.

Variable $Z = \frac{X - 65}{5}$ has the standard normal distribution

$x = 75$ gives $z = \frac{75 - 65}{5} = 2$.

Evaluate $P(B) = P(C) = P(D) = P(X < 75) = P\left(\frac{X - 65}{5} < \frac{75 - 65}{5}\right) = P(Z < 2) = 0.9772$ (refer to statistical tables)

$$P(\bar{A}) = P(B \cap C \cap D) = P(B)P(C)P(D) = P(B)^3 = 0.9772^3 \approx 0.9931$$

Law of complement $P(A) = 1 - P(\bar{A}) \approx 1 - 0.9931 = 0.0069$.

Answer: 0.0069.

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