Answer to Question #174119 in Statistics and Probability for Denisse Bisuña

Question

find the point estimate of the population parameter μ, and the standard deviation for each of the following sets of data. show your computations.

1. scores in a long test in science:

78 75 86 82 70 85 83 86

80 92 82 85 80 80 84 86

90 88 90 78 83 90 86 84

75 85 77 88 85 90 85 83

83 86 83 84 86 92 85 80

76 88 79 84 80 88 80 88

1. Lengths (in centimeters) of seedlings in a plant box:

8 8.6 12 10 8 10.5 8 10.6

8.6 10.5 7.4 6.4 12.2 6.5 12 6.8

7.5 8 11 8.5 9.5 12 11.5 12.5

10.4 7 6.8 7 7 10.5 7 7

7 8.3 7 13.5 12.5 7 7 12.5 10 6.8 10.2 6 6.5 10.3 6.8 6.8

Answer 1

1.Mean:

$\mu=\frac{\sum x_i}{N}$

$\mu=\frac{2\cdot78+2\cdot75+6\cdot86+2\cdot82+70+6\cdot85+4\cdot83+6\cdot80+2\cdot92+4\cdot84+4\cdot90+5\cdot88+77+76+79}{48}=\frac{3930}{48}=$

$=81.875\approx 82$

Standard deviation:

$\sigma=\sqrt{\frac{\sum(x_i-\mu)^2}{N}}$

$\sigma=\sqrt{\frac{2\cdot4^2+2\cdot7^2+6\cdot4^2+12^2+6\cdot3^2+4+6\cdot2^2+2\cdot10^2+4\cdot2^2+4\cdot8^2+5\cdot6^2+5^2+6^2+3^2}{48}}=4.95$

2.

$\mu=\frac{5\cdot8+2\cdot8.6+3\cdot12+2\cdot10+3\cdot10.5+10.6+7.4+6.4+12.2+2\cdot6.5+5\cdot6.8+7.5+11+8.5+9.5+11.5+3\cdot12.5}{48}+$

$+\frac{10.4+9\cdot7+8.3+13.5+10.2+6+10.3}{48}=\frac{435.5}{48}=9.07\approx9.1$

$\sigma=\sqrt{\frac{5\cdot1.1^2+2\cdot0.5^2+3\cdot2.9^2+2\cdot0.9^2+3\cdot1.4^2+1.5^2+1.7^2+2.7^2+3.1^2+2\cdot2.6^2+5\cdot2.3^2+1.6^2+1.9^2+0.6^2+0.4^2+}{48}}$

$\sqrt{\frac{+2.4^2+3\cdot3.4^2+1.3^2+9\cdot2.1^2+0.8^2+4.4^2+1.1^2+3.1^2+1.2^2}{48}}=6.675$