The length of a movie file that is available for streaming...

The length of a movie file that is available for streaming is modeled using the normal distribution shown below. The mean of the distribution is 189.4 min and the standard deviation is 20.1 min. In the figure, V is a number along the axis and is under the highest part of the curve. And, U and W are numbers along the axis that are each the same distance away from V. Use the empirical rule to choose the best value for the percentage of the area under the curve that is shaded, and find the values of U, V, and W Percentage of total area shaded: (Choose one) 120 140 160 U 180 V 200 W 220 240 260 Time (in min) U = V = W =

Transcript

00:01 Hi, i'm david and i'm here to help you answer your question.

00:04 Now let me bring up your question here.

00:06 In this question here we're going to use the empirical rule and let me remind you about the empirical rule.

00:14 So suppose that we have the normal distribution and then the middle one will be the mean.

00:25 Now suppose that we move to the right by one standard deviation, another one standard deviation, one more standard deviation.

00:34 We do the same thing on the level, we will minus one standard deviation, minus one standard deviation, minus one standard deviations.

00:43 And the empirical rule says that within the one standard deviation, it will take up to the 68%.

00:49 Within the two standard deviation, it will take up to the 95%.

00:56 And within the three standard deviations, it will take up to the 99 .7%.

01:02 Now in this question here, we have the length.

01:04 On the movie 5 it will follow by the normal with the mean now it will equal to the 189 .4 so the middle v here it will be the mean and it will equal to the 189 .4 minute.

01:20 And then we are given the standard division equal to the 20 .1.

01:25 So here percentage of total area shaded here i don't know what option you have here but i will do the one with the 68 % first.

01:34 For the 68 % we will have this one will be one standard deviation on the left.

01:41 This one will be one standard division on the right.

01:44 So therefore, the value here, it will use the 68%.

01:50 It will be by the formula will be the mu minus one standard division...

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