The mode of a grouped frequency distribution is 75 and the modal class is 65-80. The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8. Find the frequency of the modal class​

Answers 2

[tex]\large\underline{\sf{Solution-}}[/tex]

We know,

Mode of the continuous series is given by

[tex]{\boxed{ \boxed{\sf{Mode = l + \bigg(\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2} \bigg) \times h }}}} \\ \\ [/tex]

where,

l is lower limit of modal class.

[tex] \sf{f_1} [/tex] is frequency of modal class

[tex] \sf{f_0} [/tex] is frequency of class preceding modal class

[tex] \sf{f_2} [/tex] is frequency of class succeeding modal class

h is class height.

Now, from given data we concluded that

[tex]\rm \: f_0 = 6 \\ [/tex]

[tex]\rm \: f_1 = f \\ [/tex]

[tex]\rm \: f_2 = 8 \\ [/tex]

[tex]\rm \: l = 65 \\ [/tex]

[tex]\rm \: h = 15 \\ [/tex]

[tex]\rm \: Mode = 75 \\ [/tex]

So, on substituting the values in above formula, we get

[tex]\rm \: 75 = 65 + \bigg[\dfrac{f - 6}{2 \times f - 6 - 8} \bigg] \times 15\\ [/tex]

[tex]\sf \: 75 - 65 = \dfrac{f - 6}{2f - 14} \times 15 \\ \\ [/tex]

[tex]\sf \: 10 = \dfrac{f - 6}{2f - 14} \times 15 \\ \\ [/tex]

[tex]\sf \: 2 = \dfrac{f - 6}{2f - 14} \times 3 \\ \\ [/tex]

[tex]\sf \: 2(2f - 14) = 3(f - 6) \\ \\ [/tex]

[tex]\sf \: 4f - 28 = 3f -18 \\ \\ [/tex]

[tex]\sf \: 4f - 3f = 28 -18 \\ \\ [/tex]

[tex]\bf\implies \:f = 10 \\ \\ [/tex]

[tex]\rule{190pt}{2pt} \\ [/tex]

Additional Information :-

1. Mean using Direct Method

[tex]\boxed{ \rm{ \:Mean = \dfrac{ \sum f_i x_i}{ \sum f_i} \: }} \\ \\ [/tex]

2. Mean using Short Cut Method

[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i d_i}{ \sum f_i} \: }} \\ \\ [/tex]

3. Mean using Step Deviation Method

[tex]\boxed{ \rm{ \:Mean =A + \dfrac{ \sum f_i u_i}{ \sum f_i} \times h \: }} \\ \\ [/tex]

Answer:

Given :-
  • The mode of a grouped frequency distribution is 75 and the modal class is 65 - 80.
  • The frequency of the class preceding the modal class is 6 and the frequency of the class succeeding the modal class is 8.
To Find :-
  • What is frequency of the modal class.
Formula Used :-

[tex]\clubsuit[/tex] Mode Formula :-

[tex]\bigstar \: \: \sf\boxed{\bold{Mode =\: l + \bigg[\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg] \times h}}\: \: \: \bigstar\\[/tex]

where,

  • l = Lower Limit of the Modal Class
  • h = Class Size of the Class Interval
  • f₁ = Frequency of the Modal Class
  • f₀ = Frequency of the Class Preceding the Modal Class
  • f₂ = Frequency of the Class Succeeding the Modal Class
Solution :-

Let,

[tex]\mapsto \bf Frequency_{(Modal\: Class)} =\: f_1\\[/tex]

Given :

  • Mode = 75
  • Lower limit = 65
  • Class Size = 15
  • Frequency of the class preceding = 6
  • Frequency of the class succeeding = 8

According to the question by using the formula we get,

[tex]\implies \sf\boxed{\bold{Mode =\: l + \bigg[\dfrac{f_1 - f_0}{2f_1 - f_0 - f_2}\bigg] \times h}}\\[/tex]

[tex]\implies \sf 75 =\: 65 + \bigg[\dfrac{f_1 - 6}{2f_1 - 6 - 8}\bigg] \times 15\\[/tex]

[tex]\implies \sf 75 =\: 65 + \bigg[\dfrac{f_1 - 6}{2f_1 - 14}\bigg] \times 15\\[/tex]

[tex]\implies \sf 75 =\: 65 + \dfrac{f_1 - 6}{2f_1 - 14} \times 15\\[/tex]

[tex]\implies \sf 75 =\: 65 + \dfrac{15(f_1 - 6)}{2f_1 - 14}\\[/tex]

[tex]\implies \sf 75 =\: 65 + \dfrac{15f_1 - 90}{2f_1 - 14}\\[/tex]

[tex]\implies \sf 75 - 65 =\: \dfrac{15f_1 - 90}{2f_1 - 14}[/tex]

[tex]\implies \sf 10 =\: \dfrac{15f_1 - 90}{2f_1 - 14}[/tex]

By doing cross multiplication we get,

[tex]\implies \sf 15f_1 - 90 =\: 10(2f_1 - 14)[/tex]

[tex]\implies \sf 15f_1 - 90 =\: 20f_1 - 140[/tex]

[tex]\implies \sf 15f_1 - 20f_1 =\: - 140 + 90[/tex]

[tex]\implies \sf {\cancel{-}} 5f_1 =\: {\cancel{-}} 50[/tex][tex]\implies \sf 5f_1 =\: 50[/tex]

[tex]\implies \sf f_1 =\: \dfrac{\cancel{50}}{\cancel{5}}[/tex]

[tex]\implies \sf f_1 =\: \dfrac{10}{1}[/tex]

[tex]\implies \sf\bold{f_1 =\: 10}[/tex]

Hence, the required frequency of the modal class is :

[tex]\dag[/tex] Frequency Of Modal Class :

[tex]\dashrightarrow \sf Frequency_{(Modal\: Class)} =\: f_1\\[/tex]

[tex]\dashrightarrow \sf\bold{\underline{Frequency_{(Modal\: Class)} =\: 10}}\\[/tex]

[tex]\therefore[/tex] The frequency of the modal class is 10 .

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Other related questions that may help you

If the median of the following data is 33 then, find the value of m Class Frequency 0-15 17 15-30 35 30-45 40 45-60 18 60-75 m 75-90 2​

Answer:Is it correct

Step-by-step explanation:

answer img
answer img

Step-by-step explanation:

Median = l + [ (n/2 - c)/f ] × h

here, l = lower class

         n = total number of observation

         c = cumulative frequency

          f = frequency

         h = class size

 Class interval       Frequency        Cumulative Frequency

      0-15                         17                                17

      15-30                      35                               52

      30-45                      40                              92

      45-60                      18                               110

       60-75                      m                               110+m

       75-90                       2                               112+m

    33 = 30 + [ (112 + m)/2 - 52 ]/40 × 15

    3 =  (112 + m - 104)/80 × 15

    1/5 = (8 + m)/80

     16 = 8 + m

     m = 8

   

13. In the picture given below, one can see a rectangular in-ground swimming pool installed by a family in their backyard. There is a concrete sidewalk around the pool of width x m. The outside edges of the sidewalk measure 7 m and 12 m. The area of the pool is 36 sq. m. (a) Based on the information given above, form a quadratic equation in terms of x.b) Find the width of the sidewalk around the pool. ​

Answer:

(a)

[tex]2 {x}^{2} + 19x + 24 = 0[/tex]

(b)

[tex]x = \frac{3}{2} [/tex]

Step-by-step explanation:

Refer to the attachment↑↑

Hope it helps...

answer img
word usage write sentences of your own using the following word and phrasal verbs 1) detour2) formal3) solitary4) take off5) give up​
  • Detour -

The main road was blocked because of which we had to make 3 km detour.

  • Formal -

During the meeting with CEO, Rohan behaved in formal manner only.

  • Solitary -

She likes to write novels in solitary.

  • Take off -

The new Aeroplane's takeoff was very natural without any disturbances.

  • Give up -

Even after going through such a big accident, she has decided not to give up her preperation.

Answer:

1. We took a detour to the petrol pump before going to the restaurant.

2. The company sent a formal invitation to all the employees and clients of the stakeholders.

3. It was a solitary verse of the poem that made the judges give them full marks.

4. The flight was gonna take off at the Amsterdam airport.

5. She was taught by her parents, never to give up.

Explanation:

A conjunction is indeed a letter that joins words, phrases, or clauses together. The English language has many conjunctions, but some of the more popular ones include and, for, but, since, for, if, or when.

The set of rules governing the word and sentence constructions in the English language is known as English grammar.

The tense structure refers to the order in which the paragraph's components appear. Verbs, auxiliary verbs, subjects, objects, etc. are all topics of discussion. The way a sentence is put together tells the reader / listener if it is in the past, present, or future tense. The set of grammatical principles governing the English language are known as English grammar. This encompasses the organisation of individual words, sentences, clauses, phrases, and texts as a whole.

#SPJ2

two concentric circle are of radii 4cm and 3cm find the length of the chord of the larger circle which touches the smaller circle​

[tex]\large\underline{\sf{Solution-}}[/tex]

Let assume that C(O, r) and C(O, R) be two concentric circles with centre O and such that R = 4 cm and r = 3 cm.

Let further assume that AB be a chord of larger circle touches the small circle at C.

Construction :- Join OC

We know, radius and tangent are perpendicular to each other.

So, OC is perpendicular to AB.

Now, as OC is perpendicular to AB and we know, perpendicular drawn from centre bisects the chord.

So, AC = CB

Now, In right-angle triangle OAC

By using Pythagoras Theorem, we have

[tex]\sf \: {OA}^{2} = {OC}^{2} + {AC}^{2} \\ \\ [/tex]

[tex]\sf \: {R}^{2} = {r}^{2} + {AC}^{2} \\ \\ [/tex]

[tex]\sf \: {4}^{2} = {3}^{2} + {AC}^{2} \\ \\ [/tex]

[tex]\sf \: 16 = 9 + {AC}^{2} \\ \\ [/tex]

[tex]\sf \: {AC}^{2} = 16 - 9 \\ \\ [/tex]

[tex]\sf \: {AC}^{2} = 7 \\ \\ [/tex]

[tex]\bf\implies \:AC = \sqrt{7} \: cm \\ \\ [/tex]

Now,

[tex]\bf\implies \:AB = 2AC = 2\sqrt{7} \: cm \\ \\ [/tex]

answer img

Answer:

The length of the chord is [tex]2\sqrt{7}[/tex].

Step-by-step explanation:

According to the question , two concentric circle are of radii 4cm and 3cm.

We need to find the length of the chord of the larger circle which touches the smaller circle​.

If we can draw the circles according to the question we can see that the joining line of the centre and the touching point is the radius of the small circle and also it is perpendicular to the chord and if we join any one side point of the chord with the centre than we can find that it is the radius of the larger circle.

So now we can see that it is a right angle triangle. length of half of the chord = [tex]\sqrt{4^{2} - 3^{2} } = \sqrt{7}[/tex]

So the length of the chord is [tex]2\sqrt{7}[/tex].

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