Solution to Mr. Ybarola conducted an examination on his subject. He has determined that the mean score … - Sikademy
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Archangel Macsika

Mr. Ybarola conducted an examination on his subject. He has determined that the mean score of all those who had taken the test is 89 and its standard deviation is 4. 1. Determine the 𝑧-score of Karlie who scored 94 in the test. 2. What is the raw score of Sayrille, whose 𝑧-score is -0.12? 3. If a student is picked at random, what is the probability that his score lie between 75 and 85? 4. A student is picked at random. What is the probability that his score is above 93?

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Here's the Solution to this Question

1.z(94) = ( 94 - 89)/ 4 = 1.25 which is the required solution


2.Given the z score, we define

-0.12 = ( x- 89)/ 4

solving the above, we get x- 89 = -0.48

Thus x = (-0.48 + 89) = 88.52 which is the raw score.


3.We define the following

z(75) = (75-89)/4 = -3.5

z(85) =(85-89)/4 = -1

we define p(-3.5<z<-1), using the normal tables, we obtain

(0.15866) - (0.00023) = 0.15843 which is the required solution.

4.define z(93) = (93-89)/4 = 1

the probability that his score is above 93 = 1 - p(z<1)

using normal table, we obtain that p(z<1) =0.84134

Thus 1- p(z<1) = 1 - 0.84134 = 0.15866 which is the required solution.


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