Class 10

Math

All topics

Statistics

In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequency x and y.$RunsScoredNo.ofBatsmen 2500−35005 3500−4500x 4500−5500y 5500−650012 6500−75006 7500−85002 $

Runs scored | Number of Bats men | Cumulative frequency |

$2500−3500$ | $5$ | $5$ |

$3500−4500$ | $x$ | $5+x$ |

$4500−5500$ | $y$ | $5+x+y$ |

$5500−6500$ | $12$ | $17+x+y$ |

$6500−7500$ | $6$ | $23+x+y$ |

$7500−8500$ | $2$ | $25+x+y$ |

Now, $N=∑f_{i}=60$

$⇒60=25+x+y$

$⇒x+y=35$ .....$(1)$

Since the median is given to be $5000$, it lies in the class $4500−5500$

$⇒l=4500$,$h=1000$,$f=y$,$F=5+x$ and $N=60$

median$=l+f2N −F ×h$

$⇒5000=4500+y260 −(5+x) ×1000$

$⇒500=y30−5−x ×1000$

$⇒500y=(25−x)×1000$

$⇒2y =25−x$

$⇒y=50−2x$

$⇒2x+y=50$ ........$(2)$

From $(1)$ and $(2)$ we get

$2x+(35−x)=50$

$⇒x=15$

Put $x=15$ in $(1)$, we get $y=20$

Hence, $x=15$ and $y=20$