Question

12. The mean and standard deviation of an examination in which grades 70 and 88 corresponds to
Z-scores of -0.6 and 1.4, respectively are *
(1 Point)
O i = 75.4, o = 9
Oi = 9,0 = 75.4
O I = 75,0 = 9.4
O = 9.4, 0 = 75​

Answer 1

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Answer 2

Answer:

(A)The mean and standard deviation of data respectively are 75.4 and 9

Step-by-step explanation:

Given that the grades of the examination 70 and 88 correspond to the Z scores of -0.6 and 1.4.

We are required to find the Mean and standard deviation for the following information.

Let mean of the data be

$\mu$

And the standard deviation of the data be

$\sigma$

The Z score is defined by the relation as given below

$Z=\frac{Marks-Mean}{Standard \: devation} \\ \\ = > Z = \frac{x - \mu}{\sigma}$

For 70 marks,the z score is -0.6 therefore,

$- 0.6 = \frac{70- \mu}{\sigma} \\ = > 70 = \mu - 0.6\sigma - > (1)$

Similarly for 88 marks,the z score is 1.4.Therefore,

$1.4 = \frac{88 - \mu}{\sigma} \\ = > 88 = \mu + 1.4\sigma - - > (2)$

Subtracting equation 1 from equation 2,we get

$(2) - (1) = > 2\sigma = 18 \\ = > \sigma = 9$

Hence the standard deviation is 9.Substituting this value in equation 1,we get

$70 = \mu - 0.6(9) \\ = > \mu = 75.4$

Therefore,The mean and standard deviation of data respectively are 75.4 and 9

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