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.

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

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

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

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