### Mean and Standard Deviation

Hi everyone. We only had one period of math. Today we went over

That is what the 'curve' should look like. Although, the ends of the curve probably shouldn't be touching the bottom like that..I had a hard time drawing it so I hope you know what I meant to leave out.

There were two questions we did on the board.

This list shows the lengths in millimeters, of 52 arrowheads.

a) Calculate the mean and standard deviation.

b) Determine the lengths of arrowheads, one st. dev. below and one st. dev. above the mean.

c) How many arrowheads are within one st. dev. of the mean?

d) What percent of the arrowhead are within one st. dev. of the mean length?

a) First of all you put the data given into one list. Then go to [stat], arrow over to [calc], and choose 1: [1-Var Stats], [enter] then 2nd function L1. You can get the mean and standard deviation from that information.

x¯ = 25.4423 σ = 5.6207

b) Now before you calculate one st. dev. below and above the mean, it's best to store those numbers first. To do this you go to [vars], then choose 5:[statistics]. Option 2 represents the mean and option 4 represents standard deviation. To store mean; option 2, store it and set it to [alpha] M. To store standard deviation; option 4, store it and set it to [alpha] S.

One st. dev. below : x¯ - σ = 19.8216

One st. dev. above : x¯ + σ = 31.0631

c) Just count the numbers between 19.8216 and 31.0631.

37

d) 37/52 = .7115

So the percentage is 71.15%

We also repeated (b) and (d) for 2 st. dev. below and above the mean, as well as 3 st. dev. below and above the mean.

Below; x¯- 2σ = 14.2008 Above; x¯+ 2σ = 36.6838 %; 50/52 = 96.15%

Below; x¯- 3σ = 8.5801 Above; x¯+ 3σ = 42.3046 %; 52/52 = 100%

The table shows the weights (in pounds) of newborn infants.

The same 4 questions are asked. The only difference is you put the Mean of Interval list into L1 and # of Newborn Infants into L2. From there on out, it's basically the same procedure. The only difference I can think of is when you go to [1-Var stats], press enter, L1, then comma and then L2.

x¯ = 7.4 σ = 1.5798

Below; x¯- σ = 5.8201 Above; x¯ + σ = 8.9798 %; 81/125 = 64.8%

Below; x¯- 2σ = 4.2402 Above; x¯+ 2σ = 10.5597 %; 117/125 = 93.6%

Below; x¯- 3σ = 2.6603 Above; x¯+ 3σ = 12.1396 %; 125/125 = 100%

That was it for that class =) The homework for this weekend is on page 116, questions 1-9. The scribe for Monday will be: Patrick.

**normal distribution;**such that 68% of the values are within 1 standard deviation away from the mean, about 95% of the values are within two standard deviations and about 99.7% are within 3 standard deviations.That is what the 'curve' should look like. Although, the ends of the curve probably shouldn't be touching the bottom like that..I had a hard time drawing it so I hope you know what I meant to leave out.

There were two questions we did on the board.

**Here's the first one:**This list shows the lengths in millimeters, of 52 arrowheads.

16 16 17 17 18 18 18 18 19 20 20 21 21

21 22 22 22 23 23 23 24 24 25 25 25 26

26 26 26 27 27 27 27 27 28 28 28 28 29

30 30 30 30 30 30 31 33 33 34 35 39 40

a) Calculate the mean and standard deviation.

b) Determine the lengths of arrowheads, one st. dev. below and one st. dev. above the mean.

c) How many arrowheads are within one st. dev. of the mean?

d) What percent of the arrowhead are within one st. dev. of the mean length?

*Solutions:*a) First of all you put the data given into one list. Then go to [stat], arrow over to [calc], and choose 1: [1-Var Stats], [enter] then 2nd function L1. You can get the mean and standard deviation from that information.

x¯ = 25.4423 σ = 5.6207

b) Now before you calculate one st. dev. below and above the mean, it's best to store those numbers first. To do this you go to [vars], then choose 5:[statistics]. Option 2 represents the mean and option 4 represents standard deviation. To store mean; option 2, store it and set it to [alpha] M. To store standard deviation; option 4, store it and set it to [alpha] S.

One st. dev. below : x¯ - σ = 19.8216

One st. dev. above : x¯ + σ = 31.0631

c) Just count the numbers between 19.8216 and 31.0631.

37

d) 37/52 = .7115

So the percentage is 71.15%

We also repeated (b) and (d) for 2 st. dev. below and above the mean, as well as 3 st. dev. below and above the mean.

*2 st. dev:*Below; x¯- 2σ = 14.2008 Above; x¯+ 2σ = 36.6838 %; 50/52 = 96.15%

*3 st. dev:*Below; x¯- 3σ = 8.5801 Above; x¯+ 3σ = 42.3046 %; 52/52 = 100%

**Here's the second question:**The table shows the weights (in pounds) of newborn infants.

Weight Interval Mean of Interval # of Newborn Infants

3.5 - 4.5 4 4

4.5 - 5.5 5 11

5.5 - 6.5 6 1

6.5 - 7.5 7 33

7.5 - 8.5 8 29

8.5 - 9.5 9 17

9.5 - 10.5 10 8

10.5 - 11.5 11 4

Total 125

The same 4 questions are asked. The only difference is you put the Mean of Interval list into L1 and # of Newborn Infants into L2. From there on out, it's basically the same procedure. The only difference I can think of is when you go to [1-Var stats], press enter, L1, then comma and then L2.

*So here are the answers:*x¯ = 7.4 σ = 1.5798

*1 st. dev:*Below; x¯- σ = 5.8201 Above; x¯ + σ = 8.9798 %; 81/125 = 64.8%

*2 st. dev:*Below; x¯- 2σ = 4.2402 Above; x¯+ 2σ = 10.5597 %; 117/125 = 93.6%

*3 st. dev:*Below; x¯- 3σ = 2.6603 Above; x¯+ 3σ = 12.1396 %; 125/125 = 100%

That was it for that class =) The homework for this weekend is on page 116, questions 1-9. The scribe for Monday will be: Patrick.

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