Asurvey found that womens heights are normally distributed...

Asurvey found that women's heights are normally distributed with mean 63.3 in and standard deviation 3.9 in: The survey also found that men's heights are normally distributed with mean 67.4 in. and standard deviation 3.1 in. Most of the live characters employed at an amusement park have height requirements of minimum of 56 in. and maximum of 62 in. Complete parts (a) and (b) below: Find the percentage of men meeting the height requirement: What does the result suggest about the genders of the people who are employed as characters at the amusement park? The percentage of men who meet the height requirement is 4.08 % (Round to two decimal places as needed ) Since most men do not meet the height requirement, it is likely that most of the characters are women. If the height requirements are changed to exclude only the tallest 50% of men and the shortest 5% of men, what are the new height requirements? The new height requirements are a minimum of _ in. and maximum of _ in. (Round to one decimal place as needed )

Transcript

00:01 So in this question we are looking at normal distributions, so i'm going to start by drawing one.

00:08 The total area under this curve is 1 or 100%, and it is symmetric.

00:14 So 50 % below the mean, 50 % above the mean.

00:18 In part a, we're looking at the requirements for live characters at an amusement park.

00:25 So we're looking at the percentage of men who meet that height requirement.

00:29 So the parameters will be heights of men.

00:33 We have mean mu is 67 .4, standard deviation sigma is 3.

00:40 And we want to know what percentage fall within between 56, which is below the mean, and 62, which is also below the mean.

00:50 So this area here is what we're looking for, between 56 and 62.

00:57 This percentage of men meet the height requirements for these characters.

01:02 So the first step is to find the corresponding z scores.

01:06 Z is x minus mu over sigma.

01:10 It tells you how many standard deviations away from a mean a value is.

01:16 So you want z of 56 and of 62.

01:21 So for example, 62 minus 67 .4, minus 5 .4 divided by 3.

01:28 And i'll just leave that in exact form.

01:32 Actually, now it's fine.

01:33 It's only minus 1 .8, but it was going to be an ugly decimal.

01:40 For 56, 56 minus 67 .4, divided by 3 is minus 3 .8.

01:47 Okay, so our cut -off points are 1 .8 and 3 .8 standard deviations below the mean.

01:54 Now i need to turn them into area under the curve.

01:57 So you could use a z score table, a graphical calculator, or statistical software, like excel or r.

02:03 Anything with the normal functions baked into it.

02:07 There are two such functions.

02:09 They are the standard normal and the cumulative normal.

02:12 We can use either one, as long as we use them correctly.

02:15 Standard gives us the area between x and mu.

02:21 Cumulative gives you the area to the left.

02:27 So cumulative, you would put in minus 3 .8 to get this red area, put in 1 .8 minus 1 .8 to get red plus green.

02:34 And then red plus green minus red, leaves green.

02:38 I'm going to use standard in case you're using the z score table, since those are usually the standard type.

02:44 If i put in 1 .8, i get this area between this value and the mean.

02:50 And the standard table doesn't have negative scores, you just put in positive, because if you imagine a positive z score of 1 .8, this area and this area are the same.

03:01 So it can't tell which one you're looking at.

03:03 If i put in 3 .8, i get red plus green, and then i'm just the same...

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