Engineers must consider the breadths of male heads when...

Engineers must consider the breadths of male heads when designing helmets. The company researchers have determined that the population of potential clientele have head breadths that are normally distributed with a mean of 6.4-in and a standard deviation of 1.2-in. In what range would you expect to find the middle 95% of most head breadths? Between [ ] and [ ]. If you were to draw samples of size 34 from this population, in what range would you expect to find the middle 95% of most averages for the breadths of male heads in the sample? Between [ ] and [ ]. Enter your answers as numbers. Your answers should be accurate to 2 decimal places. This problem uses the count-by-thirds empirical rule.

Transcript

00:01 Male headbreadths are said to be normally distributed with a mean of 6 .4 inches and a standard deviation of 1 .2 inches.

00:08 And we were asked in what range would we expect to find the middle 95 % of most headbreadths.

00:14 So this graph represents the normal distribution of headbreadths.

00:17 We have 6 .4 inches in the center, standard deviation of 1 .2 inches.

00:22 So there exist two values.

00:24 Let's call it x2 and x1, such that 95 % of the distribution of head breaths lies within this range.

00:37 So the area under the curve between these two valleys is 0 .95.

00:41 Therefore the area under the curve outside of this range is 1 minus 0 .95, which is 0 .05.

00:48 And due to symmetry, it's half of 0 .05 in each tail.

00:55 So we can make the probability statement that the probability that a randomly selected head head breadth is at most x sub 1, it's 0 .025, and the probability that a randomly selected head breadth is at most x sub 2 is 0 .025 plus 0 .95, which is 0 .975.

01:19 We can solve this using the inverse normal distribution function in excel, which for a normal random variable allows us to provide as an argument the cumulative probability and it returns the value of the variable that corresponds to that cumulative probability.

01:40 So we use the norm .inv function that's the inverse normal distribution function.

01:45 The first argument is the cumulative probability so for x sub 1 it's 0 .25 then we enter the mean of the distribution, standard deviation deviation of the distribution, hit enter, we get approximately 5 .59.

02:03 And we can do the same thing for x sub 2, the cumulative probability is 0 .975.

02:08 So all we have to do is change the first argument to 0 .975, and we get 8 .75.

02:20 So 95 % of headbreadths lie between these two values.

02:25 And then we're asked for a random sample of 34 heads, between, within what range would what do you expect to find? the middle 95 % of the averages for samples of this size.

02:40 So samples of size 34 will have some sample mean which would vary from random sample to random sample of size 34.

02:47 This is called the sampling distribution of sample means.

02:50 Because the population is normal, this means that random sample is drawn from this population to have a distribution that is also normal.

02:59 The mean of the sampling distribution is equal to the population mean.

03:03 That's 6 .4.

03:04 And the standard deviation of sampling distribution of sample means is the population standard deviation over the square root of the sample size...

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