Based on an analysis of sample data, an article proposed the pdf f(x)=0.65e−0.65(x−1) when x≥1 as a model for the distribution of X= time (sec) spent at the median line. (a) What is the probability that waiting time is at most 4 sec? More than 4 sec? at most 4 sec P(X≤4)=? more than 4 sec P(X&gt;4)=? (b) What is the probability that waiting time is between 3 and 7 sec?

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Based on an analysis of sample data, an article proposed the pdf f(x)=0.65e−0.65(x−1) when x≥1 as a model for the distribution of X= time (sec) spent at the median line.  (a) What is the probability that waiting time is at most 4 sec? More than 4 sec?  at most 4 sec P(X≤4)=?  more than 4 sec P(X&amp;gt;4)=?  (b) What is the probability that waiting time is between 3 and 7 sec?

Oung1985 · Accepted Answer

Step 1 The pdf of X is given below: f(x)={0.65e−0.65(x−1),x≥10,otherwise a) The probability that waiting time is at most 4 sec is given below: P(x≤4)=∫140.65e−0.65(x−1)dx =[−e−0.65(x−1)]14 =−e14−0.65(x−1) =−e−0.65(4−1)−(e−0.65(1−1)) =1−0.1423 =0.8577 The probability that waiting time is more than 4 sec is given by: P(x&amp;gt;4)=1−P(x≤4) =1−0.8577 =0.1423 Step 2 b) The probability that waiting time is at most 4 sec is given below: P(x≤4)=∫370.65e−0.65(x−1)dx =[−e−0.65(x−1)]37 =−e−0.65(7−1)−(−e−0.65(3−1)) =0.2725−0.0202 =0.2523 Therefore, the indicated probability is 0.2523.

Based on an analysis of sample data, an article proposed the pdf f(

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