Did I solve the problem correctly?

[nenaquetuhace]

The following temperatures were recorded in Pasadena for a week in April
87,85,80,78,83,86,98

78
80,83,85,86,87
90

Mean = 589 / 7 = 84.14 is 84
Median - n + 1 / 1/2 = 7 + 1= 8 / 2 = 4
Standard Deviation - normally distributed
Interpret Standard Deviation - there is more data in the middle

 

[chrisr]

Check back,
was the last temperature 98 or 90?
if 98 then just add 8 to your total.
Then get the average again.

Median is the middle value, which you can see when the numbers are arranged in increasing order.
Try that again.

Standard deviation: You must calculate this, then comment about the actual value.
1. Calculate the 7 deviations 87-mean, 85-mean, up to 98-mean (or 90-mean depending on the last temperature).
2. Square all the deviations.
3. Add up all the squares.
4. Divide by n=7.
5. Get the square root of your answer from step 4.

A large value for standard deviation would imply values that differ significantly from the mean.
The temperatures you have do not differ much from the mean, hence your standard deviation is small.

 

[nenaquetuhace]

okay the mean would be 589 / 7 = 84.14 or 84

589 x2 = 346921 then 78 x 2 , 80x2,83x2,85x2,86x2,87x2,90x2 = 49663

49663 - 346921 / 7-1 = 103 - 6 then 17.16 = 17.2 sqr = 4.1 which is the standard deviation?

 

[chrisr]

No,

you must first calculate
87-84.14 = 2.86
85-84.14 = 0.86

and so on for each temperature.

You are subtracting the mean value from each temperature,
that's if the last temperature was 90 instead of 98.

When you have calculated those 7 "deviations from the mean", you square the 7 answers

(2.86)(2.86) = (2.86)[sup:34gr8591]2[/sup:34gr8591]
(0.86)(0.86) = (0.86)[sup:34gr8591]2[/sup:34gr8591]

and the same for the other 5 deviations.

Next you add up all those answers..... (2.85)[sup:34gr8591]2[/sup:34gr8591] + (0.86)[sup:34gr8591]2[/sup:34gr8591] +.....the other 5 squares.

Then, after adding them up, you divide by 7 to get an average square.
Finally get the square root of the answer.
That final answer is the standard deviation.

See how far you can go with that.

Did you find the median? You should have got 85.