Question 1188972: Suppose that a doorway being constructed is to be used by a class of people
whose heights are normally distributed with mean 70 cms. and standard deviation 3
cms. How long may the doorway be without causing more than 25% of the people
to bump their heads? If the height of the doorway is fixed at 76 cms., how many
persons out of 5000 are expected to bump their heads? (Given Φ(2) = 0.4762).
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! mean = 70 cm.
standard deviation = 3 cm.
for no more than 25% of the people to bump their heads, the height of the door needs to be at least 72.023 cm.
if the height of the doorway is fixed at 76 cm, the proportion of people who will bump their heads becomes 0.0228.
.0228 * 5000 = 114.
that's the number who are expected to bump their heads out of 5000 if the height of the door is set at 76 cm.
i do not understand what (Given Φ(2) = 0.4762) means.
given the mean and standard deviation provided, my answers are as shown above.
let me know if you have any questions or concerns.
theo
if i had used z-scores, the answer would have been the same.
for example:
the area of .25 to the right of the z-score leads to a z-score of .6744897495.
using the z-score formula of z = (x - m) / s, i get:
.6744897495 = (x - 70) / 3
solve for x to get:
x = .6744897495 * 3 + 70 = 72.023.
same as i got using the calculator.
also, using the z-score formula of z = (x - m) / s, i get:
z = (76 - 70) / 3 = z-score of 2.
area to the right of a z-score of 2 is equal to .0228.
i now see where your (Given Φ(2) = 0.4762) came from, but the area to the right of it is .0228 and the area to the left of it is .9772.
there is a form of the z-score table that just uses half the normal distribution.
in that case, the area to the left would be shown as .5 - .0228 = .4772.
if that type of table were used, the area to the left would have to have .5 added to it which would get .9772 to the left.
given that (Given Φ(2) = 0.4762), assuming it's from the half table, then you would add .5 to it to get .9762 which is pretty close to .9772 which is what i got.
if you send me the link to the table that you are using, i can verify that's what you have.
i don't really need to, becausre i'm quite confident i'm right, but it would be good to confirm, if you can send me the link to the table you are using.
if you need me to explain to you how the half table works, i can do that as well.
fyi, as far as i can tell, the half tables aren't used very much anymore, but they still do exist.
theo
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