I have a set of 19 notes, all $5 or $20, worth $215 in total. of these notes, nine are in my wallet. what is the possible range
2 answers:
You might be interested in
Answer:
a) 'X' be the random variable in discrete distribution
b) <em> The value of y = 0.2</em>
<em>c) The mean value or Expectation value</em>
<em>E (X) = 1.25</em>
<em>d) </em>
<em>The variance σ² of the discrete distribution function is</em>
<em>Variance ( V(x)= 1.2875</em>
Step-by-step explanation:
<u><em>Step(i):-</em></u>
Given data
x : 0 1 2 3 4
P(x) : 0.30 0.35 y 0.10 0.05
a)
Let 'X' be the random variable in discrete distribution
Given data is discrete distribution
<em>i) If the numbers </em> for all values of 'i'
ii) ∑P(x) = 1
Given data <em> </em> for all values of 'i'
∑P(x) = 1
0.30+0.35+y+0.10+0.05 =1
y + 0.8 = 1
y = 1 -0.8 = 0.2
<em>b) The value of y = 0.2</em>
<u><em>Step(ii):-</em></u>
<u><em>Expectation:</em></u>
Given data
x : 0 1 2 3 4
P(x) : 0.30 0.35 0.2 0.10 0.05
<em>The mean value or Expectation value</em>
<em>E (x) = ∑ x p ( X = x)</em>
<em> = 0 × 0.30 + 1 × 0.35 + 2 × 0.2 + 3 × 0.10 + 4 × 0.05</em>
<em> = 0 + 0.35 + 0.4 + 0.30 + 0.2</em>
<em> = 1.25</em>
<u><em>Variance of X</em></u>
The variance σ² of the discrete distribution function is defined by
σ² = ∑ x² p(x=x) - μ²
= 0× 0.30 + 1² × 0.35 + 2²× 0.2 +3²× 0.10 + 4²× 0.05 - (1.25)²
= 0 + 0.35 + 0.8 + 0.9 + 0.8 - 1.5625
= 1.2875
<u><em>conclusion:-</em></u>
<em>The mean value or Expectation value = 1.25</em>
<em>The variance σ² of the discrete distribution function</em>
<em> V(X) = 1.2875</em>