AP Statistics 4.8 Mean and Standard Deviation of Random Variables- MCQs - Exam Style Questions
▶️ Answer/Explanation
Mean (given): \(\mu=E[X]=1.6\).
Compute \(E[X^2]=\sum x^2P(X=x)=\dfrac{0^2(1)+1^2(12)+2^2(4)+3^2(1)+4^2(1)+5^2(1)}{20}\).
Numerator \(=0+12+16+9+16+25=78\Rightarrow E[X^2]=\dfrac{78}{20}=3.9\).
Variance: \(\operatorname{Var}(X)=E[X^2]-\mu^2=3.9-(1.6)^2=3.9-2.56=1.34\).
Standard deviation: \(\sigma_X=\sqrt{1.34}\approx1.16\).
✅ Answer: (C) \(\sigma_X\approx1.16\)
Compute \(E[X^2]=\sum x^2P(X=x)=\dfrac{0^2(1)+1^2(12)+2^2(4)+3^2(1)+4^2(1)+5^2(1)}{20}\).
Numerator \(=0+12+16+9+16+25=78\Rightarrow E[X^2]=\dfrac{78}{20}=3.9\).
Variance: \(\operatorname{Var}(X)=E[X^2]-\mu^2=3.9-(1.6)^2=3.9-2.56=1.34\).
Standard deviation: \(\sigma_X=\sqrt{1.34}\approx1.16\).
✅ Answer: (C) \(\sigma_X\approx1.16\)
▶️ Answer/Explanation
Detailed solution
The mean (expected value) of a discrete random variable is \(E(X)=\sum x_i p_i\).
\(E(X)=1(0.56)+2(0.28)+3(0.08)+4(0.06)+5(0.02)\)
\(=0.56+0.56+0.24+0.24+0.10=1.70\).
✅ Answer: (C)