Question

refer to the accompanying table, which describes results from groups of 8 births from 8 different sets of parents. the random variable x represents the number of girls among 8 children. find the mean and standard deviation for the number of girls in 8 births. click the icon to view the data table. the mean is μ = 4.0 girl(s) (round to one decimal place as needed.) the standard deviation is σ = girl(s) (round to one decimal place as needed.) table of numbers of girls and probabilities number of girls x p(x) 0 0.005 1 0.031 2 0.111 3 0.220 4 0.252 5 0.226 6 0.112 7 0.031 8 0.003

Explanation:

Step1: Recall variance formula

The formula for the variance $\sigma^{2}$ of a discrete - random variable is $\sigma^{2}=\sum(x - \mu)^{2}P(x)$, where $x$ is the value of the random variable, $\mu$ is the mean, and $P(x)$ is the probability of $x$. We know $\mu = 4.0$.

Step2: Calculate $(x-\mu)^{2}P(x)$ for each $x$

For $x = 0$: $(0 - 4)^{2}\times0.005=16\times0.005 = 0.08$.
For $x = 1$: $(1 - 4)^{2}\times0.031 = 9\times0.031=0.279$.
For $x = 2$: $(2 - 4)^{2}\times0.111 = 4\times0.111 = 0.444$.
For $x = 3$: $(3 - 4)^{2}\times0.220=1\times0.220 = 0.220$.
For $x = 4$: $(4 - 4)^{2}\times0.252 = 0\times0.252=0$.
For $x = 5$: $(5 - 4)^{2}\times0.226 = 1\times0.226 = 0.226$.
For $x = 6$: $(6 - 4)^{2}\times0.112 = 4\times0.112 = 0.448$.
For $x = 7$: $(7 - 4)^{2}\times0.031 = 9\times0.031 = 0.279$.
For $x = 8$: $(8 - 4)^{2}\times0.003=16\times0.003 = 0.048$.

Step3: Sum up the values

$\sigma^{2}=0.08 + 0.279+0.444 + 0.220+0+0.226+0.448+0.279+0.048=1.922$.

Step4: Calculate the standard - deviation

The standard deviation $\sigma=\sqrt{\sigma^{2}}=\sqrt{1.922}\approx1.4$.

Answer:

$1.4$