The equation is as follows:
Temperature in F = 32+1.8*Temperature in C.

The mean daily temperature for October was 32C with standard deviation 1.5.
1. Determine the mean daily temperature for October in F. Explain.
2. Determine the standard deviation of daily maximum temperatures, measured in F, for October. Explain.

Thanks.

2. 1. 89.6 F. It's simple substitution. Replace C with 32 and you get this equation.

F = 32+1.8*32

2. I think I know this one as well. It's just taking the 1.5 C standard deviation and converting it to Fahrenheit. I don't think you need to actually understand what standard deviation means or how it's calculated.

So you take that same formula and apply it:

F = 32+1.8*1.5 = 34.7

Hope this helps and is correct...

Originally Posted by wkeej

The equation is as follows:
Temperature in F = 32+1.8*Temperature in C.

The mean daily temperature for October was 32C with standard deviation 1.5.
1. Determine the mean daily temperature for October in F. Explain.
2. Determine the standard deviation of daily maximum temperatures, measured in F, for October. Explain.

Thanks.

3. Thanks. I will let my niece know.

4. Originally Posted by BenLin1234
1. 89.6 F. It's simple substitution. Replace C with 32 and you get this equation.

F = 32+1.8*32

2. I think I know this one as well. It's just taking the 1.5 C standard deviation and converting it to Fahrenheit. I don't think you need to actually understand what standard deviation means or how it's calculated.

So you take that same formula and apply it:

F = 32+1.8*1.5 = 34.7

Hope this helps and is correct...
Interesting, my niece after so long, informed me that the second answer is wrong. The teacher gave the answer as:
sd in F = 1.8*1.5 = 2.7.

Both she and I are lost. The teacher refuses to explain to her/other classmates asking them to find out on their own. Any idea on the answer?

5. I have no idea when it comes to using standard deviation in calculating temperature. But using common sense I think BenLin1234's calculation is right.

6. I'm sorry wkeej. I don't know why the answer is wrong either assuming the teacher is right. But we can back calculate the formula the teacher is using by using the formula for conversion from Celcius to Farenheit - 9/5(C+32).

So the formula the teacher is using is:

9/5(1.5 +32-32). I don't know why they are subtracting 32, but the mean is 32. So perhaps there's significance with that. This makes no sense to me and I frankly think it's wrong or some information is missing. Hope this helps.

7. I am just guessing here:

Standard deviation is basically the variation from mean, i.e. it is not absolute value. Therefore, the second question is basically (9/5) X 1.5 = 2.7. In other words, if the mean is 89.6°F, this value may vary as much as 2.7°.

EDIT: clarification, I did not mean 89.6±2.7°F. SD is a measurement of how good the spread is, i.e. the higher the number, the higher the variation among the raw data. Since I am not a statistician, I am afraid I am not able to explain it too well. Sorry ...

8. Originally Posted by foxs
I am just guessing here:

Standard deviation is basically the variation from mean, i.e. it is not absolute value. Therefore, the second question is basically (9/5) X 1.5 = 2.7. In other words, if the mean is 89.6°F, this value may vary as much as 2.7°.

EDIT: clarification, I did not mean 89.6±2.7°F. SD is a measurement of how good the spread is, i.e. the higher the number, the higher the variation among the raw data. Since I am not a statistician, I am afraid I am not able to explain it too well. Sorry ...
I'll take a wild stab at this:

One standard deviation indicates that roughly 68% of the time you will fall into the range specified.

Given the condition: The mean daily temperature for October was 32C with standard deviation 1.5.

This would mean 68% of the time the temperature would be within 30.5C and 33.5C.

This translates to 68% of the time the temperature would be within 86.9F and 92.3F.

Since you know the mean is 89.6F, the deviation from the mean is 2.7F (e.g. 92.3F-89.6F).

How's that?

9. Originally Posted by darkcser
I'll take a wild stab at this:

One standard deviation indicates that roughly 68% of the time you will fall into the range specified.

Given the condition: The mean daily temperature for October was 32C with standard deviation 1.5.

This would mean 68% of the time the temperature would be within 30.5C and 33.5C.

This translates to 68% of the time the temperature would be within 86.9F and 92.3F.

Since you know the mean is 89.6F, the deviation from the mean is 2.7F (e.g. 92.3F-89.6F).

How's that?
Thanks darkcer. That looks right. If I do a funny looks test on my answer, it seems wrong. 1.5 C doesn't translate into a 34.7 F when we're talking about a difference between the 2 numbers.

10. Thanks for all the help. Appreciated.

11. Originally Posted by wkeej
Interesting, my niece after so long, informed me that the second answer is wrong. The teacher gave the answer as:
sd in F = 1.8*1.5 = 2.7.

Both she and I are lost. The teacher refuses to explain to her/other classmates asking them to find out on their own. Any idea on the answer?
The question is about the properties of standard deviations. The relevant properties from the Wiki page are:

stdev(X+c)=stdev(X)
stdev(cX)=|c|stdev(X)

So in the conversion formula, adding 32 does not change the standard deviation, but the multiplication by 1.8 will scale the standard deviation by the same amount.

12. Another question:
A coach bought some T-shirts and shorts for \$256. 1/3 of them were T-shirts. The total amount he spent on the T-shirts was \$32 more than the total amount he spent on the shorts. A T-shirt cost \$22 more than a pair of shorts. Find the cost of a T-shirt.

13. T-shirt \$36, Shorts \$14, the coach bought 4 T-shirts and 8 pairs of shorts.

14. In case you want to know: let x = number of T-shirts, y = number of shorts, T = cost of one T-shirt, S = cost of a pair of shorts.
xT + yS = 256
xT = yS + 32
y = 2x
T = S + 22
Four unknowns, four equations.

15. Thanks, foxs. Looks correct.

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