Question

six children: how many females

a) how many trials were performed?
b) what is the experimental probability of having two females in
a family of six children?
c) according to this experiment, what is the average number of
females in a family with six children? how can you tell?

Explanation:

Step1: Calculate total trials

Total trials = sum of all frequencies.
$$1 + 4+6 + 3+3=17$$

Step2: Calculate experimental probability for two females

Experimental probability \(P=\frac{\text{Frequency of two females}}{\text{Total trials}}\).
Frequency of two females is \(4\), total trials \(n = 17\). So \(P=\frac{4}{17}\)

Step3: Calculate average number of females

Use the formula \(\bar{x}=\frac{\sum_{i = 0}^{5}(x_i\times f_i)}{\sum_{i=0}^{5}f_i}\)
\(x_0 = 0,f_0=1;x_1 = 1,f_1 = 0;x_2=2,f_2 = 4;x_3 = 3,f_3=6;x_4 = 4,f_4=3;x_5=5,f_5 = 3\)
\(\sum_{i = 0}^{5}(x_i\times f_i)=0\times1+1\times0 + 2\times4+3\times6+4\times3+5\times3\)
\(=0 + 0+8 + 18+12 + 15=53\)
\(\sum_{i=0}^{5}f_i=17\)
\(\bar{x}=\frac{53}{17}\approx3.12\)
The average is approximately \(3.12\) because we use the weighted - average formula (sum of (number of females \(\times\) their frequency) divided by total trials)

Answer:

a) \(17\)
b) \(\frac{4}{17}\)
c) Approximately \(3.12\) (using the weighted - average formula \(\bar{x}=\frac{\sum_{i = 0}^{5}(x_i\times f_i)}{\sum_{i=0}^{5}f_i}\))