A survey found that womens heights are normally distributed...

A survey found that womens heights are normally distributed with mean 63.3 in and standard deviation 3.5 in. The survey also found that mens heights are normally distributed with mean 68.3 in and standard deviation 3.3. Most of the live characters employed at an amusement park have height requirements of a minimum of 56 in and a max of 62 in. 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 a max of __in (Round one decimal place)

Transcript

00:01 There is a normal distribution given in this question and the given mean value.

00:05 Let me just take this note, which is denoted by the symbol mu, that is 63 .3 .3.

00:11 And the standard deviation, which is given as 3 .5.

00:17 Let's say x be the random variable for this normal distribution.

00:20 It has 63 .3 and 3 .5.

00:23 So the next thing, the survey also found that the manned heights are normally distributed.

00:29 It.

00:30 Okay.

00:34 And the given values are 63 .3 and standard divisions 3 .3.

00:41 So when we look at the last part of the question, it is being asked.

00:45 The new height requirements are a minimum of and the maximum of for the men.

00:51 So we have to just take the values which are related with men.

00:56 So actually we don't need this part because these are the information normal distribution for the woman and what about for the men so it is given that the mean value for the man which is 63 .3 and the standard deviation which is 3 .3 let's say x be the random variable for this normal distribution because it is given in the question that this is a normal distribution and so what we have to do we have to exclude only the tallest 50%, tallest 50%, and the shortest, shortest 5%.

01:42 So what we have to do, we have to just exclude this part.

01:47 So in order to just understand this question very well, maybe we can just draw a picture.

01:54 So the first thing that i will do for this question, which is just drawing a normal distribution.

02:02 So the mean value for the man, which is 68 .3.

02:07 And the tallest 50%, so the tallest 50%, which is covering all the value here, because of the 68 .3, which is the mean value.

02:19 So this divides the whole area, just two equal parts.

02:24 So this area is, which is 50 % or 0 .0.

02:33 0 .5 here and the next the shortest 5%.

02:39 So let's say there should be some x -1 value here and the shortest 5 % which is covering all this shaded region.

02:47 This is 5 % which is 0 .05.

02:52 So for the tallest 50 % and the shortest 5 % we have to exclude these two parts.

03:00 So which region we need we need just only this region.

03:03 This blue...

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