Given
In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 22.4 and a standard deviation of 5.1.
To find:
(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 19.
(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 17.7 and 27.1.
(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 33.1.
(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.
Step-by-step explanation:
It is given that,
In a recent year, the scores for the reading portion of a test were normally distributed, with a mean of 22.4 and a standard deviation of 5.1.
That implies,
a) Consider x=19.
Then,
![\begin{gathered} z=(x-\mu)/(\sigma) \\ =(19-22.4)/(5.1) \\ =-0.6667 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/b116t1gse7xo70wkz5393e95wz8se4xnv0.png)
Therefore,
The probability of a student scoring less than 19 is,
![\begin{gathered} P(x<19)=P(z<-0.6667) \\ =0.25249 \\ =0.2525 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/80ojk91rs8cl8fsmsj1yrwiyzjvzcekgoh.png)
Hence, the probability of a student scoring less than 19 is 0.2525.
b) Consider x=17.7 and 27.1.
Then,
![\begin{gathered} P(17.7<strong>Hence, the probability of a student scoring between 17.7 and 27.1 is 0.6433.</strong><p></p><p>c) Consider x=33.1.</p><p>Then,</p>[tex]\begin{gathered} P(x>33.1)=P(z>(33.1-22.4)/(5.1)) \\ =P(z>2.0980) \\ =1-P(z<2.0980) \\ =1-0.9821 \\ =0.0180 \end{gathered}]()
Hence, the probability of a student scoring more than 33.1 is 0.0180.
d) Since, a probability of 5%, or 0.05, is considered unusual.
Then, the unusual event is,
Option B),
The event in part (c) is unusual because its probability is less than 0.05.