Question

Researchers have created every possible "knockout" line in yeast. Each line has exactly one gene deleted and all the other genes present (Steinmetz et al. 2002). The growth rate - how fast the number of cells increases per hour - of each of these yeast lines has also been measured, expressed as a multiple of the growth rate of the wild type that has all the genes present. In other words, a growth rate greater than 1 means that a given knockout line grows faster than the wild type, whereas a growth rate less than 1 means it grows more slowly. Below is the growth rate of a random sample of knockout lines:

0.8, 0.98, 0.72, 1, 0.82, 0.63, 0.63, 0.75, 1.02, 0.97, 0.86

What is the standard deviation of growth rate this sample of yeast lines (answer to 3 decimals)?

Answer 1

The standard deviation of the growth rates for the given sample of yeast knockout lines is 0.148 when rounded to three decimal places.

Step-by-step explanation:

To calculate the standard deviation of the growth rates of the yeast knockout lines, we first need to compute the mean (average) of the given data. Then, following the steps, we find the variance by calculating the difference between each value and the mean, squaring those differences, and finding their average. Finally, we take the square root of the variance to find the standard deviation.

1. Calculate the mean (average) of the sample data.
2. Subtract the mean from each data point and square the result.
3. Find the average of these squared differences, which gives us the variance.
4. Take the square root of the variance to get the standard deviation.

Here are the calculations using the given growth rates:

Mean (average) = (0.8 + 0.98 + 0.72 + 1 + 0.82 + 0.63 + 0.63 + 0.75 + 1.02 + 0.97 + 0.86) / 11 = 0.836

`Variance = [(0.8 - 0.836)^2 + (0.98 - 0.836)^2 + (0.72 - 0.836)^2 + (1 - 0.836)^2 + (0.82 - 0.836)^2 + (0.63 - 0.836)^2 + (0.63 - 0.836)^2 + (0.75 - 0.836)^2 + (1.02 - 0.836)^2 + (0.97 - 0.836)^2 + (0.86 - 0.836)^2] / (11 - 1)`

Variance = 0.021918

Standard Deviation = sqrt(variance) = sqrt(0.021918) ≈ 0.148

Therefore, the standard deviation of the sample growth rates, rounded to three decimal places, is 0.148.

Answer 2

Answer: 0.144

Step-by-step explanation:

Formula to find standard deviation:
`\sigma=\sqrt{\frac{\sum_(i=1)^n(x_i-\overline{x})^2}{n-1}}`

Given : The growth rate of a random sample of knockout lines:-

0.8, 0.98, 0.72, 1, 0.82, 0.63, 0.63, 0.75, 1.02, 0.97, 0.86

Here ,

`\overline{x}=(\sum_(i=1)^(10)x_i)/(n)\\\\=(0.8+0.98+0.72+1+0.82+0.63+0.63+ 0.75+1.02+ 0.97+ 0.86)/(10)\\\\=(9.18)/(10)\approx0.83`

`\sum_(i=1)^n(x_i-\overline{x})^2=(-0.03)^2+(0.15)^2+(-0.11)^2+(0.17)^2+(-0.01)^2+(-0.2)^2+(-0.2)^2+(-0.08)^2+(0.19)^2+(0.14)^2+(0.03)^2\\\\=0.2075`

Now, the standard deviation:

`\sigma=\sqrt{(0.2075)/(10)}=0.144048602909\approx0.144`

Hence, the standard deviation of growth rate this sample of yeast lines =0.144