Question

A standardized test was given to 6000 students. The scores were normally distributed with a mean of 380 and a standard deviation of 50. How many students scored between 280 and 380?

Answer #1

Here, μ = 380, σ = 50, x1 = 280 and x2 = 380. We need to compute P(280<= X <= 380). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ

z1 = (280 - 380)/50 = -2

z2 = (380 - 380)/50 = 0

Therefore, we get

P(280 <= X <= 380) = P((380 - 380)/50) <= z <= (380 -
380)/50)

= P(-2 <= z <= 0) = P(z <= 0) - P(z <= -2)

= 0.5 - 0.0228

= 0.4772

Number of students = 6000 * 0.4772 = 2863.2

i.e. 2863 (rounded to nearest integer)

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