1. A final examination for a Statistics course has a mean of 84 and a standard deviation of 4 . Find the corresponding z score for each score. $$ \begin{array}{ll} ext { a. } 96 & ext { b. } 78\end{array} $$ 2. A student scores 60 on an English test that has a mean of 54 and standard deviation of 3 and she scores 80 on a Science test with a mean of 76 and a standard deviation of 2 . On which test did she perform better? 3. The salary of junior executives in a large corporation in Ortigas area is normally distributed with a standard deviation of Php 15,600 . Cutback is pending, at which time those who earn less than Php 85,000 will be discharged. If such a cut represents a z score of -1.28 of the junior executives, what is the mean salary of the group of junior executives?

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1. A final examination for a Statistics course has a mean of 84 and a standard deviation of 4 . Find the corresponding z score for each score. $$ \begin{array}{ll}	ext { a. } 96 & 	ext { b. } 78\end{array} $$ 2. A student scores 60 on an English test that has a mean of 54 and standard deviation of 3 and she scores 80 on a Science test with a mean of 76 and a standard deviation of 2 . On which test did she perform better? 3. The salary of junior executives in a large corporation in Ortigas area is normally distributed with a standard deviation of Php 15,600 . Cutback is pending, at which time those who earn less than Php 85,000 will be discharged. If such a cut represents a z score of -1.28 of the junior executives, what is the mean salary of the group of junior executives?

UpStudy AI · Accepted Answer

1. For each score the $$ z $$ score is calculated using  
$$
z=\frac{x-\mu}{\sigma}
$$
with $$\mu=84$$ and $$\sigma=4$$.

a. For $$ x=96 $$:  
$$
z=\frac{96-84}{4}=\frac{12}{4}=3
$$

b. For $$ x=78 $$:  
$$
z=\frac{78-84}{4}=\frac{-6}{4}=-1.5
$$

2. For the English test, the $$ z $$ score is  
$$
z=\frac{60-54}{3}=\frac{6}{3}=2.
$$  
For the Science test, the $$ z $$ score is  
$$
z=\frac{80-76}{2}=\frac{4}{2}=2.
$$  
Since both $$ z $$ scores are equal, the student performed equally well on both tests relative to her peers.

3. Given that a salary of Php 85,000 corresponds to a $$ z $$ score of $$-1.28$$ and the standard deviation is Php 15,600, we use  
$$
z=\frac{x-\mu}{\sigma}
$$  
with $$ x=85000 $$. Solving for the mean $$\mu$$:
$$
-1.28=\frac{85000-\mu}{15600} \quad \Longrightarrow \quad 85000-\mu=-1.28\times15600.
$$
Calculate:
$$
-1.28\times15600=-19968.
$$
Then,
$$
85000-\mu=-19968 \quad \Longrightarrow \quad \mu=85000+19968=104968.
$$
Thus, the mean salary of the junior executives is Php 104,968.
1. **a.** The z-score for 96 is 3. **b.** The z-score for 78 is -1.5. 2. The student performed equally well on both the English and Science tests. 3. The mean salary of the junior executives is Php 104,968.

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