Question

PLEASE SHOW ME HOW TO DO THIS

A clinic took temperature readings of 250 flu patients over a weekend and discovered the temperature distribution to be Gaussian, with a mean of 102.00°F and a standard deviation of 0.5660. Use this normal error curve area table to find the following values.

(a) What is the fraction of patients expected to have a fever greater than 103.58°F?

(b) What is the fraction of patients expected to have a temperature between 101.43°F and 102.74°F?

Answer 1

what is a guassian distribution?

Answer 2

so its another term for a normal distribution ....

Answer 3

do you have the table that is mentioned?

Answer 4

yes @ amistre64

Answer 5

can you help me with this question Im really lost

Answer 6

@amistre64

Answer 7

since we have a sample size, we most likely want to find an adjusted z score to play with

Answer 8

your table is designed as a 'from the mean' area so thats the only key thing to keep in mind, the rest is just defining the adjusted z score

Answer 9

what would you say is the formula to finnd a z score?

Answer 10

z=(mean-mu)/sigma

Answer 11

where should i go from there im still confused @amistre64

Answer 12

mean is mu but youve got the right idea

they give you a value other than the mean, the mean, and the sigma values to determine a z score with

but they also give us a sample size which allows us to adjust this to a better value by multiplying z by sqrt of the sample size

Answer 13

the rest is just addinng or subtracting table look ups

Answer 14

so we need to find z first

Answer 15

yep

Answer 16

but what should i use for the mu?

Answer 17

z = (x - mean)/sigma

x is given, mean is givenn, sigma is given

adjusted z is: z sqrt(n)

Answer 18

so for a:
(103.58-102)/0.5660=2.791=z??

Answer 19

yep, now adjust it and we are ready to use the table

Answer 20

z=7.793

Answer 21

why do we need to adjust it?

Answer 22

well, each sample from a population is going to have different values associated with them; the larger the sample size, the better it conforms to the actual population parameters

Answer 23

but im reconsidering why the sample size might have been given

Answer 24

the adjustment is seen as:\[z=\frac{x-\mu}{\sigma/\sqrt{n}}\]
which is just another way to write:\[z=\frac{(x-\mu)}{1}\frac{\sqrt{n}}{\sigma}\]

Answer 25

i got 44.137

Answer 26

i dont think its right

Answer 27

my issue is that im getting an adjustment of 44 which to me seems a little absurd and i want to reconsider the problem as using the 250 sample size as a 'large enough' size as opposed to say just 25 people

Answer 28

lets use the unadjusted score lol

Answer 29

i dont think im gteting any of this

Answer 30

getting*

Answer 31

lets ask someone more attuned to this stuff like @satellite73 or @zarkon

Answer 32

if we know the z score, the rest is using table value to determine the solution

Answer 33

not sure if theyll get the tag since the site seems a bot broken at the moment

Answer 34

I need to get this question done in one hour

Answer 35

thank you for trying anyways

Answer 36

youre welcome, if they spot the tag lets hope they can steer you in the right direction.

otherwise i would ignore the adjusment and go with z=2.791 and work the tables

Answer 37

do you know how to go from there ?

Answer 38

how to work the table and getting the final answer

Answer 39

we have a mean of 102, we want to find 103 nd greater

the table gives us an area from the mean so that is the area between 102 and 103, the total area of a side is .5000 so we need to subtract the table area from .5000

Answer 40

should i use .5660-.4974= 0.0686

Answer 41

.4974 i got it from the table using z=2.8

Answer 42

no, .5660 has to do with finding the z score, the zscore is used to find the table value

the table value is then used to find the solution.

.5000 is the propabablity of 102 to infinity

the table gives us .4974 for the probability between 102 and 103

therefore, if we subtract that prob from 102 to 103, from 102 to infinity we are left with 103 to infinity, the desired solution (assuming sqrt(n) is not important in the valuation)

Answer 43

so .5000 - .4974

Answer 44

ok and that would be 0.0026, what should i do next?

Answer 45

move on to part b :)

Answer 46

is that the answer for a?

Answer 47

i hope so :)

part b is more of the same, except we have an interval with endpoints of 101.43 and 102.74

since this is an interval that includes the mean, we would be adding the z scores that the table gives us

Answer 48

i tried a and it is correct !

Answer 49

oh good :) then 250 is superficial and is just telling us we have a large enough sample size

Answer 50

so for b i got z=-1.007 and z=1.307

Answer 51

looks like we can add the table values for z=1.0 and z=1.3

Answer 52

0.3413 and 0.4032

Answer 53

added up to .7445

Answer 54

should i use .5-.7445?

Answer 55

no, we want the interval from 101.43 to 102, added to the interval from 102 to 102.74

each of which is stated by the table as is

Answer 56

since the table gives us the propbability from the mean

Answer 57

so should i subtract each value from the table from .5 individually

Answer 58

.5-0.3413 = 0.1587
.5-0.4032 = 0.0968

Answer 59

of course not

we want to know the probability in an interval from (a to b) such the mu is a part of the interval: (a to mu to b)

the table gives us values that are from mu to x therefore we want to add (mu to a) and (mu to b)

Answer 60

the other one gave us an interval that did NOT include mu, but something other that it:
(c to infinity)

well, (mu to infinity) is .5000 and the table gives us (mu to c)

therefore: (mu to infinity) - (mu to c) = (c to infinity) was our solution

Answer 61

still confused at this point

Answer 62

add the table values together ....

Answer 63

.7445

Answer 64

then?

Answer 65

then thats the answer

Answer 66

i checked it, it is right, thank you so much :)