How Do You Convert Body Temperature from Celsius to Fahrenheit in Statistics?

[Mark53]

Homework Statement

(1) Let the random variable X be the body temperature in ◦C for a randomly chosen person during waking hours. X is assumed to be a normally distributed with mean E(X) = 37.5 and standard deviation sd(X) = 0.3. Let Y be the body temperature in ◦F for a randomly chosen person during waking hours.

(a) Find E(Y ).
(b) Find the standard deviation of Y .
(c) Find the probability that the temperature of a randomly chosen person lies between 37.5 ◦C and 38.1

Homework Equations

[/B]
F=32+9(C/5)

The Attempt at a Solution

a)
F=32+9(37.5/5) = 99.5

b)

I am unsure how to find the standard deviation

When do F=32+9(0.3/5)=32.54 which doesn't look correct

C)

calculated by looking at the normal distribution graph
P(37.5<x< 38.1)=47.5%
is there a way to show this mathematically

 

[Ray Vickson]

Homework Statement

(1) Let the random variable X be the body temperature in ◦C for a randomly chosen person during waking hours. X is assumed to be a normally distributed with mean E(X) = 37.5 and standard deviation sd(X) = 0.3. Let Y be the body temperature in ◦F for a randomly chosen person during waking hours.

(a) Find E(Y ).
(b) Find the standard deviation of Y .
(c) Find the probability that the temperature of a randomly chosen person lies between 37.5 ◦C and 38.1

Homework Equations

[/B]
F=32+9(C/5)

The Attempt at a Solution

a)
F=32+9(37.5/5) = 99.5

b)

I am unsure how to find the standard deviation

When do F=32+9(0.3/5)=32.54 which doesn't look correct

C)

calculated by looking at the normal distribution graph
P(37.5<x< 38.1)=47.5%
is there a way to show this mathematically

Use
$$\text{standard deviation} = \sqrt{\text{variance}}$$
How do you get from the variance of $X$ to the variance of $Y = 32 + (9/5)X$?

 

[Mark53]

Use
$$\text{standard deviation} = \sqrt{\text{variance}}$$
How do you get from the variance of $X$ to the variance of $Y = 32 + (9/5)X$?

Use
$$\text{standard deviation} = \sqrt{\text{variance}}$$
How do you get from the variance of $X$ to the variance of $Y = 32 + (9/5)X$?

that means the variance of X would be 0.09 how would that help solving it though?

 

[Mark53]

Use
$$\text{standard deviation} = \sqrt{\text{variance}}$$
How do you get from the variance of $X$ to the variance of $Y = 32 + (9/5)X$?

would that mean the variance of Y would be 0.09(9/5)

 

[Merlin3189]

I'm sure Ray Vickson is much better at this than I, so I hesitate to query his point, but.

I come at this sort of problem from a user's point of view rather than a mathematician's. If I look at an example such as this from WikiP's article on SD:

"As a slightly more complicated real-life example, the average height for adult men in the United States is about 70 inches, with a standard deviation of around 3 inches. This means that most men (about 68%, assuming a normal distribution) have a height within 3 inches of the mean (67–73 inches) – one standard deviation – ..."

and look at what it tells me, I do not find any need to convert to variance. It tells me the average height is 70" and 68% of men have height within 3" of that.
Now if I want to know what that is in metres or even just in feet, I simple have to convert the units of the statistics. 70" is about 5.833 ft and 3" is 0.25 ft. ANY other values - such as obtained by squaring, dividing by 12 and square rooting again (whch I guess is not exactly what RV is saying?) - must be wrong, since the statistics are telling me about a physical fact. It may well be that RV is telling you to square the SD then convert by the square of (Edit: the conversion factor) then square root back again, but that is a pointless exercise, since it will not give you a different answer.

IMO your problem is in the conversion itself. You are confusing the conversion of Centigrade degrees to Fahrenheit degrees, with the conversion of values from the Centigrade Scale to the Fahrenheit Scale. The mean is a value on the scale, but the SD is a difference measured between values on a scale. Since it is HW, I'll leave it at that for now, but (Edit: I'll) keep watching.

 

[SammyS]

b)

I am unsure how to find the standard deviation

When do F=32+9(0.3/5)=32.54 which doesn't look correct

Right. That's not correct.

What Celsius temperature is 1 standard deviation above the mean ?

What Fahrenheit temperature does that correspond to ?

Now consider Merlin's reply to get the answer in a more direct way.

 

[Mark53]

Right. That's not correct.

What Celsius temperature is 1 standard deviation above the mean ?

What Fahrenheit temperature does that correspond to ?

Now consider Merlin's reply to get the answer in a more direct way.

does that mean

=32+9(37.5+0.3/5)-99.5
=0.54

 

[Merlin3189]

does that mean

=32+9(37.5+0.3/5)-99.5
=0.54

I'll let Mark say whether that means what he said, but it doesn't make much sense to me: I get 32+9(37.5+0.3/5)-90.5=279.54 ? And what it is you were calculating, I don't know.
It might help to say what you are doing, like:

What celsius temp is mean? Mean = 37.5 ^oC (given)

What Celsius temperature is 1 standard deviation above the mean ? SD = 0.3 ^oC (given) ∴Mean +1SD = 37.5 + 0.3 = 37.8 ^oC

and so on.