Question

Below, data is a histogram of interarrivals to Walmart. The mean and standard deviation of the interarrivals are both very nearly 6.5 seconds.

Interarrival Times Lacrosse Walmart 26 March 2020, 3:30pm - 3:36pm Density (1/seconds) 10 20 30 40 Interarrivals (seconds)

Estimate, using a bit of calculus, the chance a random interarrival is between 25 seconds and 35 seconds.

No credit for an answer of zero. No credit for estimating an answer based on only a visual inspection of the graphic on the previous page.

Please answer with work (will rate) thank you!

Answer 1

the given graph loole like a exponential pattern ut follow and we have mean and st. deviation both equal to 6.5 Beuend which exibibit the proposty of exponential dist. So means st. deviation = 6.5=t - data 0.15 (rate) Do Xa expc 150015) ; fixa de de,> 0 now question i r P( 252 XC 30) = 80 0.15 e 0.184 dx 2015 Misa dhe = 0.15, sto [2018]our = - [ 2015890 - é orsang =-{0.01 -0.02 1 - - [-0.61 J = 0.01 PC25 CXC 30) = 0.01 = 1% 25 -0.15) C Scanned with CamScanner