In a data analysis, the breakdown time of an insulating fluid between electrodes at 34 kv is recorded. the times, in minutes, are as follows: 0.19, 0.78, 0.96, 1.31, 2.78, 3.16, 4.15, 4.67, 4.85, 6.50, 7.35, 8.01, 8.27, 12.06, 31.75, 32.52, 33.91, 36.71, and 72.89. calculate the sample mean and sample standard deviation

Sample mean = 14.3589

Sample standard deviation = 18.8804

Step-by-step explanation:

We are given the following information:

0.19, 0.78, 0.96, 1.31, 2.78, 3.16, 4.15, 4.67, 4.85, 6.50, 7.35, 8.01, 8.27, 12.06, 31.75, 32.52, 33.91, 36.71, 72.89

Formula:

where are data points, is the mean and n is the number of observations.  

Sum of squares of differences = 200.7590696 + 184.3878117 + 179.5317906 + 170.2750275 + 134.0720222 + 125.4164222 + 104.2226064 + 93.8757011 + 90.42008005 + 61.76305373 + 49.12534321 + 40.30913268 + 37.07528005 + 5.285159001 + 302.4487116 + 329.8238326 + 382.2436589 + 499.5695537 + 3425.884122 = 6416.4883

answer: expected value of x = 1.0

step-by-step explanation:

p(x=0) = 0.25

p(x=1) = 0.50

p(x=2) = 0.25

expected value for x = 0×0.25 + 1×0.5 + 2×0.25 = 1.0

there is no question content here, but you for the free points! : d