Astandardized exam's scores are normally distributed. in a recent year, the mean test score was 1548 and the standard deviation was 319. the test scores of four students selected at random are 1990, 1280, 2240, and 1450. find the z-scores that correspond to each value and determine whether any of the values are unusual.

The Zscore value is the number of standard deviations a given score is from the mean of the distribution. The Zscore values are ; 1.386, - 0.840, 2.169 and -0.307 respectively.

None of the values are unusual.

Given the Parameters :

The test scores, X ; 1990, 1280, 2240, 1450

Recall :

For X = 1990 ;

Zscore = (1990 - 1548) / 319 = 1.386

For X = 1280 ;

Zscore = (1280 - 1548) / 319 = - 0.840

For X = 2240 ;

Zscore = (2240 - 1548) / 319 = 2.169

For X = 1450 ;

Zscore = (1450 - 1548) / 319 = - 0.307

Therefore, the corresponding Zscore values for 1990, 1280, 2240 and 1450 are 1.386, -0.840, 2.169 and -0.307 respectively

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Step-by-step explanation:

z-score is:

z = (x − μ) / σ

For the four students:

z = (1990 − 1548) / 319 = 1.39

z = (1280 − 1548) / 319 = -0.84

z = (2240 − 1548) / 319 = 2.17

z = (1450 − 1548) / 319 = -0.31

z-scores are considered significant if they are less than -2 or greater than 2.  So the third student's score is unusual.

-4

step-by-step explanation:

(-4+8)=-(4-8)=-8+4=-4

168

step-by-step explanation:

first off, we know that he drove 3.5 hours at a rate of 48 mph.

knowing that, we want to see how many miles he drove all-together.

in this case, we can just multiply 48 by 3.5.

48 * 3.5 = 168