Below is a frequency table for the number of heads when four coins are tossed 64 times.
(a) Calculate the sample mean.
(b) Find the sample variance.
(c) Find the estimated probability there will be less than three heads on a toss.
Use the definition to calculate the sample mean
\[ \begin{align}
\text{mean} &= \sum_{i=1}^n x_i \times \frac{f_i}{64} \\
&= \frac{0(5)+1(15)+2(25)+3(16)+4(3)}{64} \\
&= \frac{125}{64} \\
&= 1.95
\end{align}\ \]
Use the definition to calculate the sample variance. Set mean = m
\[ \begin{align}
s^2 &= \frac{\sum_{i=1}^n (x_i - \text{m})^2 \times x_i}{N-1} \\
&= \frac{5(0 - 1.95)^2 + 15(1 - 1.95)^2 + 25(2 - 1.95)^2}{63} \\
&+ \frac{16(3 - 1.95)^2 + 3(4 - 1.95)^2}{63} \\
&= \frac{62.86}{63} \\
&= 1.00
\end{align}\ \]
Add the frequencies for zero, one and two heads and divide by 64.
\[ \begin{align}
P(X \lt 3) &= \frac{f_0 + f_1 + f_2}{64} \\
&= \frac{45}{64} \\
&= 0.70
\end{align}\ \]
