Question

# 4) Assume that SAT Total Scores are normally distributed with a mean of 1083 and a...

4) Assume that SAT Total Scores are normally distributed with a mean of 1083 and a standard deviation of 193. Determine the following.

4a) A student who took the SAT is randomly selected. What is the probability that the student's score is more than 1170? Round to four decimal places.

4b) What percent of the SAT Total Scores are less than 1050? Round to four decimal places.

4c) Out of 400 randomly selected SAT Total Scores, about how many would you expect to be greater than 1230?

a)

Here, μ = 1083, σ = 193 and x = 1170. We need to compute P(X >= 1170). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (1170 - 1083)/193 = 0.45

Therefore,
P(X >= 1170) = P(z <= (1170 - 1083)/193)
= P(z >= 0.45)
= 1 - 0.6736 = 0.3264

b)

Here, μ = 1083, σ = 193 and x = 1050. We need to compute P(X <= 1050). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (1050 - 1083)/193 = -0.17

Therefore,
P(X <= 1050) = P(z <= (1050 - 1083)/193)
= P(z <= -0.17)
= 0.4325

c)

Here, μ = 1083, σ = 193 and x = 1230. We need to compute P(X >= 1230). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (1230 - 1083)/193 = 0.76

Therefore,
P(X >= 1230) = P(z <= (1230 - 1083)/193)
= P(z >= 0.76)
= 1 - 0.7764 = 0.2236

= 0.2236 * 400
= 89.44
approx 89