# Z Score Formula

Z-scores are expressed in terms of standard deviations from their means. Resultantly, these z-scores have a distribution with a mean of 0 and a standard deviation of 1. The formula for calculating the standard score is given below:

\[\large Z\;Score=\frac{x-\overline{x}}{\sigma}\]

Where,

x = Standardized random variable

$\large \overline{x}$ = Mean

$\large \sigma$ = Standard deviation.

**Solved Examples**

Following are the few problems based on Z score:

** Question 1: **How well did Ram perform in her English coursework compared to the other 50 students?

**Solution: **

To answer this question, can be re-phrase it as: What percentage (or number) of students scored higher than Ram and what percentage (or number) of students scored lower than Ram? First, let’s reiterate that Ram scored 70 out of 100, the mean score was 60, and the standard deviation was 15.

Score (x) | Mean (x) | Standard Deviation () | |

English Coursework | 70 | 60 |
15 |

$\large Z\; Score=\frac{x-\overline{x}}{\sigma }$

$\large =\frac{70-60}{15}$

$\large =\frac{10}{15}$

=0.6667

The z-score is 0.67 (to 2 decimal places), but now we need to work out the percentage (or number) of students that scored higher and lower than Ram.

**Question 2: **A student wrote 2 quizzes. In first quiz, he scored 80 and in other, he scored 75. The mean and standard deviation of first quiz are70 and 15 respectively, while the mean and standard deviation of second quiz are 54 and 12 respectively. The results follow normal distribution. What can you conclude about the student’s result by seeing their z scores?

**Solution:**

Calculation of student’s Z score for first quiz:

Standardized random variable x = 80

Mean $\large \overline{x}$ = 70

Population standard deviation = 15

Formula for Z score is given below:

$\large Z\; Score=\frac{x-\overline{x}}{\sigma }$

$\large =\frac{80-70}{15}$

= 0.667

Calculation of student’s Z score for second quiz:

Standardized random variable x = 75

Mean $\large \overline{x}$ = 54

Population standard deviation = 12

Formula for Z score is given below:

$\large Z\; Score=\frac{x-\overline{x}}{\sigma }$

$\large =\frac{75-54}{12}$

= 1.75

Since Z score of second quiz is better than that of first quiz, hence it is concluded that he did better in second quiz.