Question

**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?**

Answer #1

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

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