Assume that SAT scores are approximately normally distributed with a mean of 500

Question

Assume that SAT scores are approximately normally distributed with a mean of 500 and a standard deviation of 100. Be sure to make the proper continuity correction when answering each question.

A. Estimate the probability that a random person’s SAT score is greater than 600.

B. Estimate the probability that a random person’s SAT score is less than 700.

C. Estimate the probability that a random person’s SAT score is greater than 400

D. Estimate the probability that a random person’s SAT score is less than 300

E. Estimate the probability that a random person’s SAT score is between 400 and 700 inclusive.

F. Estimate the probability that a random person’s SAT score is between 300 and 400 inclusive.

G. Estimate the probability that a random person’s SAT score is exactly 500.

H. Estimate the probability that a random person’s SAT score is exactly 300.

Explanation / Answer

as the distribution is normal

mean = 500

standard deviation = 100

the formula to be used = z =(x-mean)/standard deviation

A)

For x = 600, z = (600 - 500) / 100 = 1

Hence P(x > 600) = P(z > 1) = [total area] - [area to the left of 1]

1 - [area to the left of 1]

now from the z table we will take the value of z score = 1

    = 1 - 0.8413 = 0.1587

b)

For x = 700, the z-value z = (700 - 500) / 100 = 2

Hence P(x < 700) = P(z < 2), now from the z table we will take the value of z score = 2

And that value will be the probability required.

= [area to the left of 2] = 0.9772

c)

For x = 400, z = (400 - 500) / 100 = -1

Hence P(x > 400) = P(z > -1) = [total area] - [area to the left of -1]

1 - [area to the left of -1]

now from the z table we will take the value of z score = -1

    = 1 - 0.1587 = 0.8413

d)

For x = 300, the z-value z = (300 - 500) / 100 = -2

Hence P(x < 300) = P(z < -2), now from the z table we will take the value of z score = -2

And that value will be the probability required.

= [area to the left of -2] = 0.0228

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