Exam 4: Continuous Random Variables and Probability Distributions

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The time X (minutes) for a lab assistant to prepare the equipment for a certain experiment is believed to have a uniform distribution with A = 20 and B = 30. a. Write the pdf of X and sketch its graph. b. What is the probability that preparation time exceeds 27 minutes? c. Find the preparation mean time, then calculate the probability that preparation is within 2 minutes of the mean time? d. For any a such that 20 < a < a + 2 < 30, what is the probability that preparation time is between a and a + 2 minutes?

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a. f(x)=110 for 20x30 and =0 otherwise f ( x ) = \frac { 1 } { 10 } \text { for } 20 \leq x \leq 30 \text { and } = 0 \text { otherwise } b. P(X>27)=2730110dx=.3P ( X > 27 ) = \int _ { 27 } ^ { 30 } \frac { 1 } { 10 } d x = .3 c. E(X)=2030x110dx=x220]2030=25\left. E ( X ) = \int _ { 20 } ^ { 30 } x \cdot \frac { 1 } { 10 } d x = \frac { x ^ { 2 } } { 20 } \right] _ { 20 } ^ { 30 } = 25 25 ±\pm 2 is from 23 to 27 minutes: P(23<x<27)=2327110dx=x10]2327=.40\left. P ( 23 < x < 27 ) = \int _ { 23 } ^ { 27 } \frac { 1 } { 10 } d x = \frac { x } { 10 } \right] _ { 23 } ^ { 27 } = .40 d. P(axa+2)=aa+2110dx=.2P ( a \leq x \leq a + 2 ) = \int _ { a } ^ { a + 2 } \frac { 1 } { 10 } d x = .2 since the interval has length 2.

The cdf of checkout duration X for a book on a 2-hour reserve at a college library is given by: F(x)={0x<0x240x<212xF ( x ) = \left\{ \begin{array} { l l } 0 & x < 0 \\\frac { x ^ { 2 } } { 4 } & 0 \leq x < 2 \\1 & 2 \leq x\end{array} \right. Use this cdf to compute the following: a. P(X1)P ( X \leq 1 ) b. P(.5X1)P ( .5 \leq X \leq 1 ) c. P(X>5)P ( X > 5 ) d. The median checkout duration μˉ[H int: solve .5=F(μˉ)]\bar { \mu } [ H \text { int: solve } .5 = F ( \bar { \mu } ) ] e. F(x) to obtain the density function f(x)F ^ { \prime } ( x ) \text { to obtain the density function } f ( x )

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a. P(X1)=F(1)=14=25P ( X \leq 1 ) = F ( 1 ) = \frac { 1 } { 4 } = 25 b. P(.5X1)=F(1)F(.5)=316=.1875P ( .5 \leq X \leq 1 ) = F ( 1 ) - F ( .5 ) = \frac { 3 } { 16 } = .1875 c. P(X>.5)=1P(X.5)=1F(.5)=1516=.9375P ( X > .5 ) = 1 - P ( X \leq .5 ) = 1 - F ( .5 ) = \frac { 15 } { 16 } = .9375 d. .5=F(μ~)=μ~24μ~2=2μ~=21.414.5 = F ( \tilde { \mu } ) = \frac { \tilde { \mu } ^ { 2 } } { 4 } \Rightarrow \tilde { \mu } ^ { 2 } = 2 \Rightarrow \tilde { \mu } = \sqrt { 2 } \approx 1.414 e. f(x)=F(x)=x2 for 0x<2, and =0 otherwise f ( x ) = F ^ { \prime } ( x ) = \frac { x } { 2 } \text { for } 0 \leq x < 2 , \text { and } = 0 \text { otherwise }

Let X denote the amount of time for which a book on 2-hour reserve at a college-library is checked out by a randomly selected student and suppose that X  Let X denote the amount of time for which a book on 2-hour reserve at a college-library is checked out by a randomly selected student and suppose that X    f ( x ) = \left\{ \begin{array} { l l }  .5 x & 0 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array} \right.  Calculate the following probabilities:  a.  P (\bar{ X} \leq 1 )   b.  P ( .5 \leq \bar{X} \leq 1.5 ) f(x)={.5x0x20 otherwise f ( x ) = \left\{ \begin{array} { l l } .5 x & 0 \leq x \leq 2 \\0 & \text { otherwise }\end{array} \right. Calculate the following probabilities: a. P(Xˉ1)P (\bar{ X} \leq 1 ) b. P(.5Xˉ1.5)P ( .5 \leq \bar{X} \leq 1.5 )

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a. P(X1)=1f(x)dx=015xdx=.25x2]01=.25\left. P ( X \leq 1 ) = \int _ { - \infty } ^ { 1 } f ( x ) d x = \int _ { 0 } ^ { 1 } \cdot 5 x d x = .25 x ^ { 2 } \right] _ { 0 } ^ { 1 } = .25 b. P(.5X1.5)=5155xdx=25x2]31.5=.5\left. P ( .5 \leq X \leq 1.5 ) = \int _ { 5 } ^ { 15 } \cdot 5 x d x = 25 x ^ { 2 } \right] _ { 3 } ^ { 1.5 } = .5

The lifetime X (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters α=2 and β=3\alpha = 2 \text { and } \beta = 3 Compute the following: a. E(X) and V(X) b. P(X3)P ( X \leq 3 ) c. P(1.5X6)P ( 1.5 \leq X \leq 6 )

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Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime X (in weeks) has a gamma distribution with mean of 40 and variance of 320. a. What is the probability that a transistor will last between 1 and 40 weeks? b. What is the probability that a transistor will last at most 40 weeks? Is the median of the lifetime distribution less than 40? Why or why not?

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"Time headway" in traffic flow is the elapsed time between the time that one car finishes passing a fixed point and the instant that the next car begins to pass that point. Let X = the time headway for two randomly chosen consecutive cars on a freeway during a period of heavy flow. The following pdf of X is f(x)={.15e1(x5)x50 otherwise f ( x ) = \left\{ \begin{array} { c l } .15 e ^ { - 1 ( x - 5 ) } & x \geq 5 \\0 & \text { otherwise }\end{array} \right. What is the probability that the time headway is a. At most 6 seconds? b. At least 6 seconds? c. At most 5 seconds? d. Between 5 and 6 seconds?

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Stress is applied to a 20-in. steel bar that is clamped in a fixed position at each end. Let X = distance from the left end at which the bar snaps. Suppose Y/20 has a standard beta distribution with E(Y) = 10 and V(Y) = 1007\frac { 100 } { 7 } a. What are the parameters of the relevant standard beta distribution? b. Compute P(8Y12)P ( 8 \leq Y \leq 12 ) c. Compute the probability that the bar snaps more than 2 in. from where you expect it to.

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Let X be the temperature in C- C at which a certain chemical reaction takes place, and let Y be the temperature in F- F (so Y = 1.8X + 32). a. If the median of the X distribution is μ2\mu ^ { 2 } , show that 1.8 μ~\tilde { \mu } + 32 is the median of the Y distribution. b. How is the 90th percentile of the Y distribution related to the 90th percentile of the X distribution? Verify your conjecture. c. More generally, if Y = aX + b, how is any particular percentile of the Y distribution related to the corresponding percentile of the X distribution?

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A college professor always finishes his lectures within 2 minutes after the bell rings to end the period and the end of the lecture. Let X = the time that elapses between the bell and the end of the lecture and suppose the pdf of X is f(x)={kx20x20 otherwise f ( x ) = \left\{ \begin{array} { l l } k x ^ { 2 } & 0 \leq x \leq 2 \\0 & \text { otherwise }\end{array} \right.  A college professor always finishes his lectures within 2 minutes after the bell rings to end the period and the end of the lecture. Let X = the time that elapses between the bell and the end of the lecture and suppose the pdf of X is  f ( x ) = \left\{ \begin{array} { l l }  k x ^ { 2 } & 0 \leq x \leq 2 \\ 0 & \text { otherwise } \end{array} \right.     a. Find the value of k. [Hint: Total area under the graph of f(x) is 1.] b. What is the probability that the lecture ends within 1minutes of the bell ringing? c. What is the probability that the lecture continues beyond the bell for between 60 and 90 seconds? d. What is the probability that the lecture continues for at least 90 seconds beyond the bell? a. Find the value of k. [Hint: Total area under the graph of f(x) is 1.] b. What is the probability that the lecture ends within 1minutes of the bell ringing? c. What is the probability that the lecture continues beyond the bell for between 60 and 90 seconds? d. What is the probability that the lecture continues for at least 90 seconds beyond the bell?

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Suppose the proportion X of surface area in a randomly selected quadrat that is covered by a certain plant has a standard beta distribution with α=5 and β=2\alpha = 5 \text { and } \beta = 2 a. Compute E(X) and V(X). b. Compute P(X2)P ( X \leq 2 ) c. Compute P(.2X4)P ( .2 \leq X \leq 4 ) d. What is the expected proportion of the sampling region not covered by the plant?

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Construct a normal probability plot for the following sample of observations on coating thickness for low-viscosity paint. Would you feel comfortable estimating population mean thickness using a method that assumed a normal population distribution? 83 .98 1.06 1.14 1.20 1.25 1.29 1.40 1.48 1.49 1.51 1.62 1.65 1.71 1.76 1.83

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Let X = hourly median power (in decibels) of received radio signals transmitted between two cities. It is believed that the lognormal distribution provides a reasonable probability model for X. If the parameter values are μ=3.5 and σ=1.2\mu = 3.5 \text { and } \sigma = 1.2 \text {, } calculate the following: a. The mean value and standard deviation of received power b. The probability that received power is between 50 and 250 dB c. The probability that X is less than its mean value. Why is this probability not .5?

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Let X have a binomial distribution with parameters n = 25 and p. Calculate each of the following probabilities using the normal approximation (with the continuity correction) for the cases p = .5 and .6, and compare to the exact probabilities calculated from the cumulative binomial probabilities table available in your text. a. P(10X15)P ( 10 \leq X \leq 15 ) b. P(X10)P ( X \leq 10 ) c. P(X15)P ( X \geq 15 )

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The distribution of resistance for resistors of a certain type is known to be normal, with 10% of all resistors having a resistance exceeding 10.634 ohms, and 5% having a resistance smaller than 9.7565 ohms. What are the mean value and standard deviation of the resistance distribution?

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Let X have a uniform distribution on the interval [a, b]. a. Obtain an expression for the (100p) th percentile. b. Compute E(X), V(X), and σx\sigma _ { x } . c. For n a positive integer, compute E(Xn)E \left( X ^ { n } \right) .

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Suppose only 40% of all drivers in Florida regularly wear a seatbelt. A random sample of 500 drivers is selected. What is the probability that a. Between 170 and 220 (inclusive) of the drivers in the sample regularly wear a seatbelt? b. Fewer than 175 of those in the sample regularly wear a seatbelt?

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Let X = the time between two successive arrivals at the drive-up window of a local bank. If X has an exponential distribution with λ\lambda = 1 (which is identical to a standard gamma distribution with α\alpha =1), compute the following: a. The expected time between two successive arrivals. b. The standard deviation of the time between two successive arrivals. c. P(Xˉ5)P ( \bar{X} \leq 5 ) . d. P(3Xˉ5)P ( 3 \leq \bar{X} \leq 5 ) .

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