In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of and a standard deviation of . Complete parts (a) through (d) below. (a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than . The probability of a student scoring less than is nothing. (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 between and . The probability of a student scoring between and is nothing. (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 . The probability of a student scoring more than is nothing. (Round to four decimal places as needed.) (d) Identify any unusual events. Explain your reasoning. Choose the correct answer below. A. than 0.05. B. than 0.05. C. The event in part is unusual because its probability is less than 0.05. D. The events in parts are unusual because its probabilities are less than 0.05.

The question is incomplete. Here is the complete question.

In a recent year, the socres for the reading portion of a test were normally distributed, with a mean of 23.3 and a standard deviation of 6.4. Complete parts (a) through (d) below.

(a) Find the probability that a randomly selected high school student who took the reading portion of the test has a score that is less than 18. (Round to 4 decimal places as needed.)

(b) Find a probability that a random selected high school student who took the reading portion of the test has a score that is between 19.9 and 26.7.

(c) Find a probability that a random selected high school student who took the reading portion of the test ahs a score that is more than 36.4.

(d) Identify any unusual events. Explain your reasoning.

(a) P(X<18) = 0.2033

              (b) P(19.9<X<26.7) = 0.4505

              (c) P(X>36.4) = 0.0202

               (d) Unusual event: P(X>36.4)

Step-by-step explanation: First, determine the z-score by calculating:

Then, use z-score table to determine the values.

(a) x = 18

z = -0.83

P(X<18) = P(z< -0.83)

P(X<18) = 0.2033

(b) x=19.9 and x=26.7

z = -0.67

z = 0.53

P(19.9<X<26.7) = P(z<0.53) - P(z< -0.67)

P(19.9<X<26.7) = 0.7019 - 0.2514

P(19.9<X<26.7) = 0.4505

(c) x=36.4

z = 2.05

P(X>36.4) = P(z>2.05) = 1 - P(z<2.05)

P(X>36.4) = 1 - 0.9798

P(X>36.4) = 0.0202

(d) Events are unusual if probability is less than 5% or 0.05. So, part (c) has an unusual event.