Question

In a recent​ year, the total scores for a certain standardized test were normally​ distributed, with a mean of 500 and a standard deviation of 10.5. Answer parts ​(a)dash​(d) below. ​(a) Find the probability that a randomly selected medical student who took the test had a total score that was less than 489. The probability that a randomly selected medical student who took the test had a total score that was less than 489 is . 1474. ​(Round to four decimal places as​ needed.) ​(b) Find the probability that a randomly selected medical student who took the test had a total score that was between 498 and 511. The probability that a randomly selected medical student who took the test had a total score that was between 498 and 511 is . 3922. ​(Round to four decimal places as​ needed.) ​(c) Find the probability that a randomly selected medical student who took the test had a total score that was more than 523. The probability that a randomly selected medical student who took the test had a total score that was more than 523 is nothing. ​(Round to four decimal places as​ needed.) ​(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below. A. The events in parts left parenthesis a right parenthesis and left parenthesis b right parenthesis are unusual because their probabilities are less than 0.05. B. The event in part left parenthesis a right parenthesis is unusual because its probability is less than 0.05. C. None of the events are unusual because all the probabilities are greater than 0.05. D. The event in part left parenthesis c right parenthesis is unusual because its probability is less than 0.05.  

Answer 1

µ = 500, σ = 10.5

a)

P(X < 489) =

= P( (X-µ)/σ < (489-500)/10.5 )

= P(z < -1.0476)

Using excel function:

= NORM.S.DIST(-1.0476, 1)

= 0.1474

b)

P(498 < X < 511) =

= P( (498-500)/10.5 < (X-µ)/σ < (511-500)/10.5 )

= P(-0.1905 < z < 1.0476)

= P(z < 1.0476) - P(z < -0.1905)

Using excel function:

= NORM.S.DIST(1.0476, 1) - NORM.S.DIST(-0.1905, 1)

= 0.4281

c)

P(X > 523) =

= P( (X-µ)/σ > (523-500)/10.5)

= P(z > 2.1905)

= 1 - P(z < 2.1905)

Using excel function:

= 1 - NORM.S.DIST(2.1905, 1)

= 0.0142

d)

Answer : D. The event in part (c) is unusual because its probability is less than 0.05.