MATH SOLVE

3 months ago

Q:
# 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)?

Accepted Solution

A:

Answer: 0.144Step-by-step explanation:Formula to find standard deviation: [tex]\sigma=\sqrt{\dfrac{\sum_{i=1}^n(x_i-\overline{x})^2}{n-1}}[/tex]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, 0.86Here , [tex]\overline{x}=\dfrac{\sum_{i=1}^{10}x_i}{n}\\\\=\dfrac{0.8+0.98+0.72+1+0.82+0.63+0.63+ 0.75+1.02+ 0.97+ 0.86}{10}\\\\=\dfrac{9.18}{10}\approx0.83[/tex][tex]\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[/tex]Now, the standard deviation: [tex]\sigma=\sqrt{\dfrac{0.2075}{10}}=0.144048602909\approx0.144[/tex]Hence, the standard deviation of growth rate this sample of yeast lines =0.144