Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 18: Problem 724

Mean of following frequency distribution is \(9.3\) then \(\mathrm{K}\) is $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{Xi} & 4 & 6 & 7 & \mathrm{~K} & 12 & 14 \\ \hline \mathrm{Fi} & 5 & 6 & 8 & 10 & 2 & 9 \\ \hline \end{array} $$ (a) 11 (b) 12 (c) 10 (d) 13

Short Answer

Expert verified
The value of K is 11, making the correct option (a) 11.

Step by step solution

01

Write down the given information

The frequency distribution table and mean provided are: $$ \begin{array}{|l|l|l|l|l|l|l|} \hline \mathrm{Xi} & 4 & 6 & 7 & \mathrm{~K} & 12 & 14 \\\ \hline \mathrm{Fi} & 5 & 6 & 8 & 10 & 2 & 9 \\\ \hline \end{array} $$ Mean: \(9.3\)
02

Recall the formula for calculating the mean

The formula for calculating the mean of a frequency distribution is: Mean = \(\frac{\sum (Xi \cdot Fi)}{\sum Fi}\)
03

Substitute the given information into the formula and solve for K

Plug the given information into the formula and set it equal to the mean (9.3): \(9.3 = \frac{(4 \cdot 5) + (6 \cdot 6) + (7 \cdot 8) + (K \cdot 10) + (12 \cdot 2) + (14 \cdot 9)}{5 + 6 + 8 + 10 + 2 + 9}\) Solve for K: \(9.3 \times (5 + 6 + 8 + 10 + 2 + 9) = (4 \cdot 5) + (6 \cdot 6) + (7 \cdot 8) + (K \cdot 10) + (12 \cdot 2) + (14 \cdot 9)\) Calculate the sum of the frequencies: \(9.3 \times 40 = (20) + (36) + (56) + (10K) + (24) + (126)\) Multiply the mean by the sum of the frequency: \(372 = (20) + (36) + (56) + (10K) + (24) + (126)\) Combine like terms and solve for K: \(372 - (20 + 36 + 56 + 24 + 126) = 10K\) \(110 = 10K\) Divide both sides by 10: \(K = 11\)
04

State the final answer

The value of K is 11, making the correct option (a) 11.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The median of a set of 7 distinct observations is \(10.5\) If each of the last 3 observation of the set is increased by 3 then the median of the new set \(=\) (a) in decreased by 2 (b) is two times the original median (c) remain the same as that of the original set (d) is increased by 2

A fair dice is thrown 20 Times. The probability that on the tenth throw the fourth six appear is (a) \(\left[\left\\{\left({ }^{20}{ }_{10}\right) \cdot 5^{6}\right\\} / 6^{20}\right]\) (b) \(\left[\left(120 \times 5^{7}\right) / 6^{10}\right]\) (c) \(\left[\left(84 \times 5^{6}\right) / 6^{10}\right]\) (d) None

Mean of n observations is \(\mathrm{m}\) and sum of \(\mathrm{n}-3\) observations is b then mean of remaining 3 observations is (a) \(n m+b\) (b) \([(\mathrm{nm}-\mathrm{b}) / 3]\) (c) \([(n m+b) / 3]\) (d) \(n m-b\)

The arithmetic mean of 7 consecutive integers starting with a is \(m\) then the arithmetic mean of 11 consecutive integers starting with \(a+2\) is (a) \(2 \mathrm{a}\) (b) \(2 \mathrm{~m}\) (c) \(a+4\) (d) \(m+4\)

If the mean of the set of numbers \(x_{1}, x_{2}, \ldots \ldots \ldots x_{n}\) is \(\underline{x}\) then the mean of the numbers \(x i+2 i, 1 \leq i \leq n\) is (a) \(\underline{x}+2 n\) (b) \(\underline{x}+n+1\) (c) \(\underline{x}+2\) (d) \(\underline{x}+n\)
See all solutions

Recommended explanations on English Textbooks

Graphology

Read Explanation

Essay Writing Skills

Read Explanation

Sociolinguistics

Read Explanation

Blog

Read Explanation

English Grammar

Read Explanation

Linguistic Terms

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free