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Principles of Plant Genetics and Breeding

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150 CHAPTER 9<br />

Using data in Table 9.2, the mean <strong>of</strong> the F 2 can be<br />

obtained as:<br />

X ¯ =∑X i f i /n<br />

= 2,105/193<br />

= 10.91 cm<br />

Measures <strong>of</strong> dispersion<br />

Measures <strong>of</strong> dispersion or variability concerns the<br />

degree to which values <strong>of</strong> a data set differ from their<br />

computed mean. The most commonly used measure <strong>of</strong><br />

dispersion is the mean square deviation or variance. The<br />

population variance is given by:<br />

σ 2 = [∑(X 1 −µ) 2 ]/N<br />

where σ 2 = population variance, X 1 = value <strong>of</strong> observations<br />

in the population, µ=mean <strong>of</strong> the population, <strong>and</strong><br />

N = total number <strong>of</strong> observations in the population.<br />

The sample variance is given by:<br />

s 2 = [∑(X − X ¯ ) 2 ]/(n − 1)<br />

where s 2 = sample variance, X = value <strong>of</strong> the observation<br />

in the sample, X ¯ = mean <strong>of</strong> the sample, <strong>and</strong> n = total<br />

number <strong>of</strong> observations in the sample.<br />

The computational formula is:<br />

s 2 = [∑X 2 − (∑X ¯ ) 2 /n]/(n − 1)<br />

Using the data below for number <strong>of</strong> leaves per plant:<br />

Number <strong>of</strong> leaves Total<br />

X 7 6 7 8 10 7 9 8 7 10 79 =∑X<br />

X 2 49 36 49 64 100 49 81 64 49 100 641 =∑X 2<br />

(∑X) 2 /n = 79 2 /10 = 6,241/10 = 624.1<br />

s 2 = (641 − 642.1)/9 = 16.9/9 = 1.88<br />

Variance may also be calculated from grouped data.<br />

Using the data in Table 9.2, variance may be calculated<br />

as follows:<br />

s 2 = [n∑fx 2 − (∑fx) 2 ]/n(n − 1)<br />

= [193(24,705) − (2,105) 2 ]/193(193 − 1)<br />

= (4,768,065 − 4,431,025)/37,056<br />

= 337,040/37,056<br />

= 9.10 (for F 2 , the most variable generation<br />

following a cross)<br />

Variance for the F 1 is 1.67.<br />

St<strong>and</strong>ard deviation<br />

The st<strong>and</strong>ard deviation (SD) measures the variability<br />

that indicates by how much the value in a distribution<br />

typically deviates from the mean. It is the positive square<br />

root <strong>of</strong> the population variance. The larger the value <strong>of</strong><br />

the st<strong>and</strong>ard deviation, the more the observations (data)<br />

are spread about the mean, <strong>and</strong> vice versa.<br />

The st<strong>and</strong>ard deviation <strong>of</strong> the sample is simply:<br />

s =√s 2<br />

For the number <strong>of</strong> leaves per plant example:<br />

s =√1.88<br />

= 1.37<br />

Similarly, for the seedling height data:<br />

SD <strong>of</strong> the F 2 =√9.1 = 3.02<br />

SD <strong>of</strong> the F 1 =√1.67 = 1.29<br />

Normal distribution<br />

One <strong>of</strong> the most important examples <strong>of</strong> continuous<br />

probability distribution is the normal distribution or<br />

the normal curve. It is important because it approximates<br />

many kinds <strong>of</strong> natural phenomena. If the population<br />

is normally distributed, the mean = 0.0 <strong>and</strong> the<br />

variance = 1.0. Further, a range <strong>of</strong> ±1 SD from the<br />

mean will include 68.26% <strong>of</strong> the observations, whereas<br />

a range <strong>of</strong> ±2 SD from the mean will capture most <strong>of</strong><br />

the observations (95.45%) (Figure 9.1). The shape<br />

<strong>of</strong> the curve varies depending on the nature <strong>of</strong> the<br />

population.<br />

68.3%<br />

95.4%<br />

99.7%<br />

–3σ –2σ –1σ µ +1σ +2σ +3σ<br />

Figure 9.1 Normal distribution curve.

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