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# Tutor profile: Brian D.

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Brian D.
Data Scientist at US Marine Corps (Yale '20: M.A. Statistics)
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## Questions

### Subject:Pre-Calculus

TutorMe
Question:

Solve the equation $$\log_4(4x+4)=3$$.

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Brian D.
Answer:

Since the base of the logarithm is $$4$$, then we take exponents for both sides with base $$4$$ to eliminate $$\log$$ in the equation. This leads to: $$4^{\log_4(4x+4)}=4^3$$, or $$4x+4=64$$. We solve this linear equation so that $$x=15$$.

### Subject:Calculus

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Question:

Which value is larger: $$I_1=\int_{0}^{1} x \space dx$$ or $$I_2=\int_{0}^{1} x^2 \space dx$$?

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Brian D.
Answer:

The value $$I_1$$ is larger. Solving the integral for $$I_1$$ leads to: $$\int_{0}^{1} x \space dx=\frac{x^2}{2} |_{x=0}^{x=1}=\frac{1^2}{2}-\frac{0^2}{2}=\frac{1}{2}$$. Solving the integral for $$I_2$$ leads to: $$\int_{0}^{1} x^2 \space dx=\frac{x^3}{3} |_{x=0}^{x=1}=\frac{1^3}{3}-\frac{0^3}{3}=\frac{1}{3}$$. Thus, $$\frac{1}{2}=I_1>I_2=\frac{1}{3}$$.

### Subject:Statistics

TutorMe
Question:

There exists a random variable $$X \sim \text{Binomial}(n, p)$$. Calculate the moment generating function (MGF) of the random variable $$X$$, in terms of the parameters $$n$$ and $$p$$.

Inactive
Brian D.
Answer:

From the property of Binomial distributions, we know that $$X = \sum_{i=1}^{n} Y_i$$ is the sum of $$n$$ independent and identically distributed Bernoulli random variables $$Y_1,...,Y_n$$. The formula of the moment generating function (MGF) is: $$M_X(t)=\mathbb{E}[e^{tX}]$$. So, the MGF of a Bernoulli random variable $$Y_i$$ is: $$M_{Y_i}(t)=\mathbb{E}[e^{tY_i}]=pe^t+(1-p)$$. This is because $$\mathbb{P}[Y_i=1]=p$$ and $$\mathbb{P}[Y_i=0]=1-p$$. The formula for the MGF of the sum of independent random variables is the product of each random variable's MGF. In other words, we have: $$M_{X}(t)=M_{\sum_{i=1}^{n} Y_i}(t)=\prod_{i=1}^{n} M_{Y_i}(t)=(pe^t+(1-p))^n$$.

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