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For the distribution of ages in the example in Section 1.3.1:

(a) Computej2 andj2 .

(b) Determine ∆j for each j, and use Equation 1.11 to compute the standard deviation.

(c) Use your results in (a) and (b) to check Equation 1.12.

Short Answer

Expert verified
  1. j2=441 and j2=459.57.
  2. Standard deviation is4.309,Δjfor eachjis given in the solution.
  3. The values using this method matches the values we got.

Step by step solution

01

Given data

The given equation 1.12 is,

σ=j2j2σ2=j2

02

Taking the average value of j

Here, j represents the ages of the individuals.

LetN(j)be the number of individuals with age j.

Hence,

role="math" localid="1655374067007" N(14)=1N(15)=1N(16)=3N(22)=2N(24)=2N(25)=5

Average of the values is given by,

j=jjNjjNjj=14×1+15×1+16×3+22×2+24×2+25×51+1+3+2+2+5j=14+15+48+44+48+12514j=21

03

Solving for  j2 and ⟨j⟩2

Since we just calculated the expectation value, hence squaring it for the required answer:

<j>2=212<j>2=441

Now solving for <j2>.

<j2>=jj2NjjNj<j2>=142×1+152×1+162×3+222×2+242×2+252×51+1+3+2+2+5<j2>=196+225+768+968+1152+312514<j2>=32177<j2>=459.57

04

Computing the standard deviation (b)

The standard deviation is given by the equation,

σ=<j2><j>2σ=459.57441σ=4.309

05

Making a table for Δj (b)

Difference between each j and the average value <j>is given by Δj.

Hence, Δj=jj

jΔj=j21(Δj)2Nj1474911563611652532211224392254165

06

Solving for the standard deviation from the given equation

Given equation is, σ2=(Δj)2

j2=j(Δj)2NjjNjj2=49×1+36×1+25×3+1×2+9×2+16×51+1+3+2+2+5j2=49+36+75+2+18+8014σ2=1307σ=4.309

Since the answer for σ is same, therefore the equation 1.12 is satisfied.

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Most popular questions from this chapter

A particle is represented (at time t=0) by the wave function

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Ψ(x)=0; Otherwise

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