Advanced Placement (AP), 03.02.2020 21:50 anniejordan0412

[edited] need on stats question 40 points

answer the new questions.
in advance.

[edited] need on stats question 40 points answer the new questions. in adva

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Answer from: user1234536

A — 1B — ∑[k=0..2] 15Ck(.24^k)(.76^(15-k))D — 12 and 3.1A — 0.1091B — 0.83 ...Explanation:

1. The probability is 0.52 that the officer will pull over a driver and then the expected number more. So, if x is the expected number of drivers pulled over until one is not texting, we have ...

... x = 0.52(1+x)

... 0.48x = 0.52

... x = 0.52/0.48 = 13/12 = 1 1/12 ≈ 1 . . . . matches selection A

(Comment on this result: I find it interesting that these are the odds in favor of finding a driver who texts. That is, if the probability of texting is 0.98, the odds are 49:1 that a driver will be texting, and the expected number of pull-overs is 49.)

2. The probability of at most 2 being cured is the probability of 0, 1, or 2 being cured. You need to add up those probabilities. The sum in answer selection B does that.

3. The mean of a binomial distribution is ...

... μx = np = 60·0.2 = 12

... σx = √(np(1-p)) = √(12·0.8) ≈ 3.0984

These match selection D.

4. 20C14(0.8^14)(0.2^6) = 38760·.043980·0.000064 ≈ 0.109100 . . . matches A

5. mean(x) = 0.94; mean(x^2) = 1.58, so ...

... σx = √(1.58 -0.94²) ≈ 0.83 . . . . matches selection B

Answer from: Quest

You need to figure out at which x-value does the graph has a tangent line slope of -1.remember, the derivative is the slope of the tangent line. so find the derivative function and set it to equal -1 to find the x-value that will have a tangent line slope of 0.differentiate with the chain rule. $\begin{aligned} f(x) & = (0.5)^{2x} \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot (2x)' \\ f'(x) & = 0.5^{2x} \cdot \ln(0.5) \cdot 2 \\ f'(x) & = 2\ln(0.5) \cdot 0.5^{2x} \end{aligned}$ set the derivative to equal -1 and solve for x. $\begin{aligned} -1 & = 2\ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2} & = \ln(0.5) \cdot 0.5^{2x} \\ -\frac{1}{2\ln(0.5)} & = 0.5^{2x} \end{aligned}$ take log base 10 of both sides. also recall that $\log_c(m^n) = n \log_c m$ . $\begin{aligned} \log\left[ -\frac{1}{2\ln(0.5)}\right] & = \log\left[0.5^{2x}\right] \\ \log\left[ -\frac{1}{2\ln(0.5)}\right] & = 2x \log(0.5) \\ \dfrac{\log\left[ -\frac{1}{2\ln(0.5)}\right]}{2\log(0.5)} & = x \\ x & \approx 0.2356 \end{aligned}$ so the x-coord of the point of tangency is $x \approx 0.2356$ .the y-coordinate of this point of tangency can be calculated using the original function: $\begin{aligned} f(x) & = (0.5)^{2x} \\ f(0.2356) & = (0.5)^{2 \cdot 0.2356} \\ & \approx 0.7213 \end{aligned}$ using point-slope form, the equation of the tangent line is $y - 0.7213 = -1(x-0.2356)$ the x-intercept of this tangent line is when the graph is at y=0. $\begin{aligned} 0 - 0.7213 & = -1(x-0.2356)\\ - 0.7213 & = -(x-0.2356) \\ 0.7213 & = x-0.2356\\ x & = 0.9569 \end{aligned}$ this rounds off to $x & = 0.957$ .choice c.

Answer from: Quest

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[edited] need on stats question 40 points

answer the new questions.
in advance.

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