Question

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. 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 19.
The probability of a student scoring less than 19 is.
(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 17.7 and 27.1.
The probability of a student scoring between 17.7 and 27.1 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 33.1.
The probability of a student scoring more than 33.1 is
3.1
is.
(Round to four decimal places as needed.)
(d) Identify any unusual events. Explain your reasoning. Choose the correct answer below.
OA. None of the events are unusual because all the probabilities are greater than 0.05.
B. The event in part (c) is unusual because its probability is less than 0.05.
C. The events in parts (a) and (b) are unusual because its probabilities are less than 0.05.
D. The event in part (a) is unusual because its probability is less than 0.05.
Help me solve this

Answer 1

Using the normal distribution, we have that:

a) The probability of a student scoring less than 19 is 0.2709 = 27.09%.

b) The probability of a student scoring between 17.7 and 27.1 is 0.6424 = 64.24%.

c) The probability of a student scoring more than 33.1 is 0.0179 = 1.79%.

d) The correct option is: B. The event in part (c) is unusual because its probability is less than 0.05.

Normal Probability Distribution

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

• The z-score measures how many standard deviations the measure is above or below the mean.

• Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 22.4, \sigma = 5.1$

Item a:

The probability is the p-value of Z when X = 19, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{19 - 22.4}{5.1}$

Z = -0.67.

Z = -0.67 has a p-value of 0.2709.

The probability of a student scoring less than 19 is 0.2709 = 27.09%.

Item b:

The probability is the p-value of Z when X = 27.1 subtracted by the p-value of Z when X = 17.7, hence:

X = 27.1:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{27.1 - 22.4}{5.1}$

Z = 0.92.

Z = 0.92 has a p-value of 0.8212.

X = 17.7:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{17.7 - 22.4}{5.1}$

Z = -0.92.

Z = -0.92 has a p-value of 0.1788.

0.8212 - 0.1788 = 0.6424.

The probability of a student scoring between 17.7 and 27.1 is 0.6424 = 64.24%.

Item c:

The probability is one subtracted by the p-value of Z when X = 33.1, hence:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33.1 - 22.4}{5.1}$

Z = 2.1.

Z = 2.1 has a p-value of 0.9821.

1 - 0.9821 = 0.0179.

The probability of a student scoring more than 33.1 is 0.0179 = 1.79%.

Item d:

Probabilities less than 0.05 are unusual, hence the correct option is:

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

More can be learned about the normal distribution at brainly.com/question/28135235

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