Chapter 15: Q15MP (page 777)

Out of 1095people, what is the probability that exactly 2 were born on Jan.1 ? Assume 365days in a year

Expert verified

The value of the function from binomial distribution is 0.224.

Step by step solution

Number of people is $1095$

Frequency distribution of the number of successful outcomes that can be achieved in a given number of trials, each with an equal chance of success.

The formula for binomial distribution states that. $f (x) = C (x, n) p^{x} q^{n - x}$

$\begin{matrix} n = 1095 \\ p = \frac{1}{365} \\ x = 2 \end{matrix}$

Substitute the values mentioned above in the formula.

$\begin{matrix} f (2) = C (1095, 2) {(\frac{1}{365})}^{2} {(\frac{364}{365})}^{1093} \\ = 0.224 \end{matrix}$

The formula for corresponding normal approximation states that . $f (x) = \frac{e^{- {(x - n p)}^{2} / 2 n p q}}{\sqrt{2 π n p q}}$

$\begin{matrix} n = 1095 \\ p = \frac{1}{365} \\ x = 2 \end{matrix}$

Substitute the values mentioned above in the formula.

$\begin{matrix} f (2) = \frac{e^{- 3} \cdot 3^{2}}{2!} \\ = 0.224 \end{matrix}$

The value of the function from binomial distribution is . $0.224$

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