In a recent year, the scores for the reading portion of a test were normally dis

Question

In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 20 3 and a standard deviation of 5 3 below (a) Find the probability that a randomly The probability of a student scoring less than 18 high school student who took the reading portion of the test has a score that is less than 18 (Round to four decimal places as needed ) (b) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is betweon 12 5 and 281 The probability of a student scoring between 12 5 and 28 1 is (Round to four decimal places as needed ) (c) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is more than 31.1 The probability of a student scoring more than 31.1 is (Round to four decimal places as needed ) (d) Identify any unusual events Explain your reasoning Choose the correct answer below A. The event in part (a) is unusual because its probability is less than 0 05 O B. None of the events are unusual because all the probabilities are greater than 0.05 O C. The event in part (c) is unusual because its probability is less than 0.05 O D. The events in parts (a) and (b) are unusual because its probabilties are less than 0 05

Explanation / Answer

Mean = 20.3

standard devition = 5.3

(a) Pr(X < 18)

Z = (18 - 20.3)/5.3 = -0.434

Pr(X < 18) = 0.3322

(b) Pr(12.5 < X < 28.1) = Pr(X < 28.1) - Pr(X < 12.5)

Z2 = (28.1 - 20.3)/5.3 = 1.4717

Z1 = (12.5 - 20.3)/ 5.3 = -1.4717

Pr(12.5 < X < 28.1) = 0.92945 - 0.07055 = 0.8589

(c) Pr(X > 31.1)

Z = (31.1 - 20.3)/ 5.3 = 2.03774

Pr(X > 31.1)= 1- Pr(X < 31.1) = 1 - 0.9792 = 0.0208

The event in (c) is unusual as its probability is less than 0.05.

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