Question

A standardized​ exam's scores are normally distributed. In a recent​ year, the mean test score was and the standard deviation was . The test scores of four students selected at random are ​, ​, ​, and . Find the​ z-scores that correspond to each value and determine whether any of the values are unusual. The​ z-score for is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for is nothing. ​(Round to two decimal places as​ needed.) Which​ values, if​ any, are​ unusual? Select the correct choice below​ and, if​ necessary, fill in the answer box within your choice. A. The unusual​ value(s) is/are nothing. ​(Use a comma to separate answers as​ needed.) B. None of the values are unusual.

Answer 1

The​ z-score for 1880 is 1.34.

The​ z-score for 1190 is -0.88.

The​ z-score for 2130 is 2.15.

The​ z-score for 1350 is -0.37.

And the z-score of 2130 is considered to be unusual.

Explanation:

The complete question is: A standardized​ exam's scores are normally distributed. In recent​ years, the mean test score was 1464 and the standard deviation was 310. The test scores of four students selected at random are ​1880, 1190​, 2130​, and 1350. Find the​ z-scores that correspond to each value and determine whether any of the values are unusual. The​ z-score for 1880 is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for 1190 is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for 2130 is nothing. ​(Round to two decimal places as​ needed.) The​ z-score for 1350 is nothing. ​(Round to two decimal places as​ needed.) Which​ values, if​ any, are​ unusual? Select the correct choice below​ and, if​ necessary, fill in the answer box within your choice. A. The unusual​ value(s) is/are nothing. ​(Use a comma to separate answers as​ needed.) B. None of the values are unusual.

We are given that the mean test score was 1464 and the standard deviation was 310.

Let X = standardized​ exam's scores

The z-score probability distribution for the normal distribution is given by;

Z =
`(X-\mu)/(\sigma)` ~ N(0,1)

where,
`\mu` = mean test score = 1464

`\sigma` = standard deviation = 310

S, X ~ Normal(
`\mu=1464, \sigma^(2) = 310^(2)`)

Now, the test scores of four students selected at random are ​1880, 1190​, 2130​, and 1350.

So, the z-score of 1880 =
`(X-\mu)/(\sigma)`

=
`(1880-1464)/(310)` = 1.34

The z-score of 1190 =
`(X-\mu)/(\sigma)`

=
`(1190-1464)/(310)` = -0.88

The z-score of 2130 =
`(X-\mu)/(\sigma)`

=
`(2130-1464)/(310)` = 2.15

The z-score of 1350 =
`(X-\mu)/(\sigma)`

=
`(1350-1464)/(310)` = -0.37

Now, the values whose z-score is less than -1.96 or higher than 1.96 are considered to be unusual.

According to our z-scores, only the z-score of 2130 is considered to be unusual as all other z-scores lie within the range of -1.96 and 1.96.