## Percentiles for grouped data

Percentiles are the values which divide whole distriution into **hundred** equal parts. They are 99 in numbers namely `$P_1, P_2, \cdots, P_{99}$`

. Here `$P_1$`

is first percentile, `$P_2$`

is second percentile and so on.

## Formula

For discrete frequency distribution, the formula for $i^{th}$ percentile is

`$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$`

where,

- $N$ is total number of observations.

For continuous frequency distribution, the formula for $i^{th}$ percentile is

`$P_i=l + \bigg(\dfrac{\frac{iN}{100} - F_<}{f}\bigg)\times h; \quad i=1,2,\cdots,99$`

where,

- $l :$ the lower limit of the $i^{th}$ percentile class
- $N=\sum f :$ total number of observations
- $f :$ frequency of the $i^{th}$ percentile class
- $F_< :$ cumulative frequency of the class previous to $i^{th}$ percentile class
- $h :$ the class width

## Example 1

A librarian keeps the records about the amount of time spent (in minutes) in a library by college students. Data is as follows:

Time spent | 30 | 32 | 35 | 38 | 40 |
---|---|---|---|---|---|

No. of students | 8 | 12 | 20 | 10 | 5 |

Calculate `$P_{15}$`

and `$P_{40}$`

.

### Solution

$x_i$ | $f_i$ | $cf$ | |
---|---|---|---|

30 | 8 | 8 | |

32 | 12 | 20 | |

35 | 20 | 40 | |

38 | 10 | 50 | |

40 | 5 | 55 | |

Total | 55 |

The formula for $i^{th}$ percentile is

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

**Fiftieth percentile $P_{15}$**

```
$$
\begin{aligned}
P_{15} &=\bigg(\dfrac{15(N)}{100}\bigg)^{th}\text{ value}\\
&= \bigg(\dfrac{15(55)}{100}\bigg)^{th}\text{ value}\\
&=\big(8.25\big)^{th}\text{ value}
\end{aligned}
$$
```

The cumulative frequency just greater than or equal to $8.25$ is $20$. The corresponding value of $X$ is the $15^{th}$ percentile. That is, $P_{15} =32$ minutes.

**Fourtieth percentile $P_{40}$**

```
$$
\begin{aligned}
P_{40} &=\bigg(\dfrac{40(N)}{100}\bigg)^{th}\text{ value}\\
&= \bigg(\dfrac{40(55)}{100}\bigg)^{th}\text{ value}\\
&=\big(22\big)^{th}\text{ value}
\end{aligned}
$$
```

The cumulative frequency just greater than or equal to $22$ is $40$. The corresponding value of $X$ is the $40^{th}$ percentile. That is, $P_{40} =35$ minutes.

## Example 2

The following table gives a frequency distribution of weight (in pounds) of 57 children at a day care center.

Weight | 10-19 | 20-29 | 30-39 | 40-49 | 50-59 | 60-69 | 70-79 |
---|---|---|---|---|---|---|---|

children | 5 | 19 | 10 | 13 | 4 | 4 | 2 |

Calculate

a. the maximum weight of lower 30 % of the children,

b. the minimum weight of upper 30 % of the children,

c. the limits for the weight of middle 40 % of the children.

### Solution

Let $X$ denote the weight of children at a day care center.

Here the classes are inclusive. To make them exclusive type subtract 0.5 from the lower limit and add 0.5 to the upper limit of each class.

Class Interval | Class Boundries | $f_i$ | $cf$ | |
---|---|---|---|---|

10-20 | 9.5-20.5 | 5 | 5 | |

20-30 | 19.5-30.5 | 19 | 24 | |

30-40 | 29.5-40.5 | 10 | 34 | |

40-50 | 39.5-50.5 | 13 | 47 | |

50-60 | 49.5-60.5 | 4 | 51 | |

60-70 | 59.5-70.5 | 4 | 55 | |

10-20 | 9.5-20.5 | 2 | 57 | |

Total | 57 |

a. The maximum weight of lower $30$ % of the children is $P_{30}$.

The formula for $i^{th}$ percentile is

$P_i =\bigg(\dfrac{i(N)}{100}\bigg)^{th}$ value, $i=1,2,\cdots, 99$

where $N$ is the total number of observations.

```
$$
\begin{aligned}
P_{30} &=\bigg(\dfrac{30(N)}{100}\bigg)^{th}\text{ value}\\
&= \bigg(\dfrac{30(57)}{100}\bigg)^{th}\text{ value}\\
&=\big(17.1\big)^{th}\text{ value}
\end{aligned}
$$
```

The cumulative frequency just greater than or equal to $17.1$ is $24$, the corresponding class $19.5-30.5$ is the $30^{th}$ percentile class.

Thus

- $l = 19.5$, the lower limit of the $30^{th}$ percentile class
- $N=57$, total number of observations
- $f =19$, frequency of the $30^{th}$ percentile class
- $F_< = 5$, cumulative frequency of the class previous to $30^{th}$ percentile class
- $h =10$, the class width

The thirtieth percentile $P_{30}$ can be computed as follows:

```
$$
\begin{aligned}
P_{30} &= l + \bigg(\frac{\frac{30(N)}{100} - F_<}{f}\bigg)\times h\\
&= 19.5 + \bigg(\frac{\frac{30*57}{100} - 5}{19}\bigg)\times 10\\
&= 19.5 + \bigg(\frac{17.1 - 5}{19}\bigg)\times 10\\
&= 19.5 + \big(0.6368\big)\times 10\\
&= 19.5 + 6.3684\\
&= 25.8684 \text{ pounds}
\end{aligned}
$$
```

The maximum weight of lower $30$ % of the children is $P_{30}= 25.8684$ pounds.

b. The minimum weight of upper $30$ % of the children is $P_{70}$.

```
$$
\begin{aligned}
P_{70} &=\bigg(\dfrac{70(N)}{100}\bigg)^{th}\text{ value}\\
&= \bigg(\dfrac{70(57)}{100}\bigg)^{th}\text{ value}\\
&=\big(39.9\big)^{th}\text{ value}
\end{aligned}
$$
```

The cumulative frequency just greater than or equal to $39.9$ is $47$, the corresponding class $39.5-50.5$ is the $70^{th}$ percentile class.

Thus

- $l = 39.5$, the lower limit of the $70^{th}$ percentile class
- $N=57$, total number of observations
- $f =13$, frequency of the $70^{th}$ percentile class
- $F_< = 34$, cumulative frequency of the class previous to $70^{th}$ percentile class
- $h =10$, the class width

The seventieth percentile $P_{70}$ can be computed as follows:

```
$$
\begin{aligned}
P_{70} &= l + \bigg(\frac{\frac{70(N)}{100} - F_<}{f}\bigg)\times h\\
&= 39.5 + \bigg(\frac{\frac{70*57}{100} - 34}{13}\bigg)\times 10\\
&= 39.5 + \bigg(\frac{39.9 - 34}{13}\bigg)\times 10\\
&= 39.5 + \big(0.4538\big)\times 10\\
&= 39.5 + 4.5385\\
&= 44.0385 \text{ pounds}
\end{aligned}
$$
```

The minimum weight of upper $30$ % of the children is $P_{30}= 44.0385$ pounds.

c. The limits for the weight of middle $40$ % of the children is `$P_{30}$`

and `$P_{70}$`

.

Thus the limits for the weight of middle 40 % of the children is `$P_{30} = 25.8684$`

pounds and `$P_{70}= 44.0385$`

pounds.