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Please assist me with the missing answers. Scores on the final exam in Ms. Meyer's statistics class follow a normal distribution, with a mean of 82 and a standard deviation of 5. 1. Ron, who took Ms Meyer's class, scored 88 on the final. Ron's z-score is + ___ 2. This means that Ron's final score is A. 1.2 points above the mean B. 1.2 standard deviation above the mean C. 1.2 points below the mean D 1.2 standard deviation below the mean

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JeanaShupp

Answer: Ron's z-score is +1.2.

B. 1.2 standard deviation above the mean .

Step-by-step explanation:

Let X deontes the scores of the final exam

Given : Ms. Meyer's statistics class follow a normal distribution with Mean =  82

standard deviation =5.1

Formula : [tex]z=\dfrac{X-Mean}{Standard \ deviation}[/tex]

Ron scored 88 on the final.

So, put x=88

[tex]z=\dfrac{88-82}{5.1}=\dfrac{6}{5.1}\approx1.2[/tex]

Ron's z-score is +1.2.

Also, 88= 86+(1.2 x 5.1) = Mean +1.2 x (standard deviation)

i.e.  This means that Ron's final score is 1.2 standard deviation above the mean .