Imagine selecting a US high school student at random. Define the random variable X = number of languages spoken by the randomly selected student. The table below gives the probability distribution of X, based on a sample of students from the US Census at School database.
What is the expected average number of languages a student knows?
Find and interpret the standard deviation for the data.
Answers
Answer:
Expected value = 1.457
Standard deviation = 0.671
Step-by-step explanation:
Given the data:
Language (X) : 1 ____ 2 ____ 3 ____ 4 ____ 5
P(X) _____0.630 _ 0.295 _0.065_0.008_0.002
Expected value :
ΣX*P(x) = (1*0.630)+(2*0.295)+(3*0.065)+(4*0.008)+(5*0.002)
= 1.457
Standard deviation : sqrt(Var(x))
Var(X) = Σx^2p(x) - E(x)
Var(X) = (1^2*0.630)+(2^2*0.295)+(3^2*0.065)+(4^2*0.008)+(5^2*0.002) - 1.457^2
= 2.573 - 1.457^2
= 0.450151
Standard deviation = sqrt(0.450151) = 0.6709329
= 0.671
Average Expected number of languages known by student, ie Expected Value = 1.457 ; Standard Deviation of Data = 0.45015
A] Expected Value represents the average, calculated by summing products of outcomes & their corresponding probabilities. E (X) = X . P (X) , where X are events & P (X) are corresponding probabilities.
E (X) = 1 (0.63) + 2 (0.295) + 3 (0.065) + 4 (0.008) + 5 (0.002) = 1.457
B] Standard Deviation denotes the measure of dispersion, spread around average. Formula = [tex]X^2 P (X) - [ E (X) ]^2[/tex]
1 (0.63) + [tex]2^2[/tex] (0.295) + [tex]3^2[/tex] (0.02,65) + [tex]4^2[/tex] (0.008) + [tex]5^2[/tex] (0.002) - [tex](1.457)^2[/tex]
= 2.573 - (1.457)^2 = 0.45015
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