1 - JustAnswer



1. Consider the data set: 70, 72, 84, 88, 90, 96, 96

(a) Find the mean and standard deviation.

(70+72+84+88+90+96+96)/7=596/7=85.14

(-15.14²+-13.14²+-1.14²+2.86²+4.86²+10.86²+10.86²)=(229.31+172.73+1.31+8.16+23.59+117.88+117.88)=670.86/6=111.81. √111.81=10.57

(b) Add 10 to each observation to create a new data set. Find the mean and standard deviation for this new data set. How are these values related to the answers in (a)?

Adding 10 just moves the mean to 95.14 without changing the distribution, so the standard deviation is the same.

(c) Multiply each observation in the original data set by 3 to create a new data set. Find the mean and standard deviation for this new data set. How are these values related to the answers in (a)?

The mean is the same, but the standard deviation is now 31.72, triple what it was.

2. A marketing researcher asked 73 adults how many times they went out to a movie in the last month. Show how to use the frequency distribution to find the standard deviation for the survey responses summarized in the table below:

x |0 |1 |2 |3 |4 |5 |7 |9 |12 |15 |20 | |f |12 |16 |12 |10 |15 |2 |1 |1 |2 |1 |1 | |12*0+1*16+2*12+3*0+4*15+5*2+7*1+6*1+12*2+15*1+20*1

0+16+24+0+60+10+7+6+24+15+20 = 182/73 = 2.49

12(-2.49)² +16(-1.49)² +12(-0.49)²+10(0.51)²+15(1.51)²+

2(2.51)²+1(4.51)²+1(6.51)²+2(9.51)²+1(12.51)²+1(17.51)²=

74.40+35.52+2.88+2.60+34.20+12.60+20.34+42.38+180.88+156.50+360.60=

922.90 / 72 = 12.82. √12.82 = 3.58

3. The following data are the price of gas at 15 gas stations (in $/gallon):

1.46 1.49 1.54 1.59 1.60 1.75 1.75 1.76 1.78 1.79 1.79 1.79 1.81 1.92 2.06

(a) Find the mean and standard deviation for the given data.

(1.46+1.49+1.54+1.59+1.60+1.75+1.75+1.76+1.78+1.79+1.79+1.79+1.81+1.92+2.06)/15

25.88/15=1.73

(-.27²+-.24²+-.19²+-.14²+.13²+.02²+.02²+.03²+.05²+.06²+.06²+.06²+.08²+.18²+.33²)/14

(.0729+.0576+.0361+.0196+.0196+.0004+.0004+.0009+.0025+.0036+.0036+.0064+.0324+.1089)/14

0.3649/14 = 0.0261. √0.0261 = 0.1616

(b) What proportion of the data lies within:

▪ one standard deviation of the mean.

▪ two standard deviations of the mean

▪ three standard deviations of the mean?

10/15 = 2/3 are within 1σ

14/15 are within 2σ

All are within 3σ

4. A manufacturer of pencils is choosing between two pencil making machines. Machine A produces pencils which have a mean length of 6 inches and a standard deviation of 0.1 inch while machine B produces pencils which have a mean length of 6 inches and a standard deviation of 0.4 inch. Which machine should the manufacturer choose and why?

With the mean being what they want, they should go with the more accurate machine, A.

5. [pic]

Orange General Car

Computer Company

Stock Price Stock Price

40 40

42 39

38 40

52 43

36 43

40 45

48 46

50 50

(40+42+38+52+36+40+48+50)/8 = 346/8 = 43.25

(-3.25²+-1.25²+-5.25²+8.75²+-7.25²+-3.25²+4.75²+6.75²)/7

(10.56+1.56+27.56+76.56+52.56+10.56+22.56+45.56)/7

247.5/7 = 35.38. √35.38 = 5.95

(40+39+40+43+43+45+46+50)/8 = 346/8 = 43.25

(-3.25²+-4.25²+-3.25²+0.25²+0.25²+1.75²+2.75²+6.75²)/7

(10.56+18.06+10.56+0.06+0.06+3.06+7.56+45.56)/7

95.5/7 = 13.64. √13.64 = 3.69

Someone who is willing to put up with more risk for the hope of a better return would be considering investing in Orange.

6. In many data collection situations, one hopes for a relatively small standard deviation. For instance, when monitoring products produced on an assembly line, one hopes for uniformity in terms of length, weight, . . . Low standard deviation implies consistency which is often a desirable quality for consumers. Can you think of any data collection situations for which a researcher might have reason to hope for a HIGH standard deviation instead of a low one?

A situation where the mean is an undesired result, such as the mean of a population is overweight, but the variance is high, so not everyone is overweight.

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