Question

For students in a certain region, scores of students on a standardized test approximately follow a normal distribution with mean ?=531.5μ=531.5 and standard deviation ?=28.1σ=28.1. In completing the parts below, you should use the normal curve area table that is included in your formula packet.

(a) What is the probability that a single randomly selected
student from among all those in region who took the exam will have
a score of 536 or higher?

ANSWER:

For parts (b) through (e), consider a random sample of 35 students
who took the test.

(b) The mean of the sampling distribution of ?¯x¯ is

(c) The standard deviation of the sampling distribution of ?¯x¯ is

(d) What z-score corresponds to the mean score of ?¯x¯ =
536?

ANSWER:

(e) What is the probability that the mean score ?¯x¯ of these 35
students will be 536 or higher?

ANSWER:

Answer #1

Solution: Given that μ = 531.5, σ = 28.1,

(a) P(X >= 536) = P(X > 536.5)

P(X > 536.5) = P((X-μ)/σ > (536.5-531.5)/28.1)

= P(Z > 0.1779)

= 0.4286

For n = 35

(b) The mean of the sampling distribution of x¯ is 531.5

(c) The standard deviation of the sampling distribution of x¯ is
4.7498

=> σx = σ/sqrt(n) = 28.1/sqrt(35) = 4.7498

(d) P(X >= 536) = P(X > 536.5)

P(X > 536.5) = P((X-μ)/(σ/sqrt(n)) >
(536.5-531.5)/(28.1/sqrt(35)))

= P(Z > 1.0527)

= 0.1469

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