AIOU Solved Assignments 1 & 2 Code 6401 Autumn & Spring 2020. Solved Assignments code 6401 General Mathematics and Statistics in Pakistan 2020. Allama iqbal open university old papers.

## Aiou Solved Assignments 2 code 6401 Autumn & Spring 2020

**Course: General Mathematics and Statistics (6401)****Level: B.Ed (2.5 Years)****Semester: Autumn & Spring 2020****AIOU Solved ASSIGNMENT No. 2**

** **

# Aiou Solved Assignments 1 & 2 Autumn & Spring 2020 code 6401

# Aiou Solved Assignments 1 & 2 code 6401 Autumn & Spring 2020

# Aiou Solved Assignments 1 & 2 code 6401

**Answer:**

**Histograms**

**Frequency Distribution Tables**

A frequency distribution table is a table that shows how often a data point or a group of data points appears in a given data set. To make a frequency distribution table, first divide the numbers over which the data ranges into intervals of equal length. Then count how many data points fall into each interval.

If there are many values, it is sometimes useful to go through all the data points in order and make a tally mark in the interval that each point falls. Then all the tally marks can be counted to see how many data points fall into each interval. The “tally system” ensures that no points will be missed.

*Example*: The following is a list of prices (in dollars) of birthday cards found in various drug stores:

1.45 | 2.20 | 0.75 | 1.23 | 1.25 |

1.25 | 3.09 | 1.99 | 2.00 | 0.78 |

1.32 | 2.25 | 3.15 | 3.85 | 0.52 |

0.99 | 1.38 | 1.75 | 1.22 | 1.75 |

Make a frequency distribution table for this data.

*We omit the units (dollars) while calculating. The values go from 0.52 to 3.85, which is roughly 0.50 to 4.00. We can divide this into 7 intervals of equal length: 0.50 – 0.99, 1.00 – 1.49, 1.50 – 1.99, 2.00 – 2.49, 2.50 – 2.99, 3.00 – 3.49, and 3.50 – 3.99. Then we can count the number of data points which fall into each interval–for example, 4 points fall into the first interval: 0.75, 0.78, 0.55, and 0.99–and make a frequency distribution table*:

Intervals (in dollars) | Frequency |

0.50 – 0.99 | 4 |

1.00 – 1.49 | 7 |

1.50 – 1.99 | 3 |

2.00 – 2.49 | 3 |

2.50 – 2.99 | |

3.00 – 3.49 | 2 |

3.50 – 3.99 | 1 |

Total | 20 |

**Frequency Polygons**

A frequency polygon is another type of frequency distribution graph. In a frequency polygon, the number of observations is marked with a single point at the midpoint of an interval. A straight line then connects each set of points. Frequency polygons make it easy to compare two or more distributions on the same set of axes. (Hennekens, 1987, p. 218) Let’s look at an example of a frequency polygon.

Notice the dotted outline of a histogram for the same data. A frequency polygon smoothes out the abrupt changes that may appear in a histogram, and is useful for demonstrating continuity in the variables being studied. In this example, the number of reported cases of influenza-like illness peaked during week 4 after the onset of illness.

Like a histogram, frequency polygons are used to display the entire frequency distribution (counts) of a continuous variable. They must be closed at both ends because the area under the curve represents all of the data. By contrast, an arithmetic-scale line graph represents a series of observed data points (counts or rates), usually over time – it simply plots data points.

**Frequency curve: **A smooth curve which corresponds to the limiting case of a histogram computed for a frequency distribution of a continuous distribution as the number of data points becomes very large.

**Less Than OGIVE:**– The less than cumulative frequencies are in ascending order. The cumulative frequency of each class is plotted against the upper limit of the class interval in this type of OGIVE and then various points are joined by straight line.

**More Than OGIVE:-** The cumulative frequencies in this type are in the descending order. The cumulative frequency of each class is plotted against the lower limit of the class interval.

Example: Marks obtained by the students of a class in statistics test are:

Marks | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |

Number of students | 4 | 8 | 18 | 15 | 5 |

Draw ‘less than’ and ‘more than’ OGIVES.

Solutions. First, the ‘less than and ‘more than’ cumulative frequencies will be calculated and the OGIVES will be drawn on the basis of these cumulative frequencies.

Calculation of Cumulative Frequencies

Marks | Frequency | ‘Less than’ Cumulative Frequency | ‘More than’ Cumulative Frequency |

0 – 10 10 – 20 20 – 30 30 – 40 40 – 50 | 4 8 18 15 5 | 4 12 30 45 50 | 50 46 38 20 5 |

6401 assignment no 2

Code 6401 assignment no. 2

Assignment no. 2 of 6401 autumn 2019

Aoa sir I need assignment no.1,2 code 6401

Asalamoalaikum sir I need BED autum 2018 1st semester assignment book code is 8611

hojygi