Question

according to a gallup poll, 69% of american adults prefer saving over spending. suppose 10 american ad are randomly chosen. use the formula for binomial probabilities

p(k)=left(\begin{array}{l}n \\ kend{array}
ight) p^{k}(1-p)^{n - k}

to find the probability that exactly 7 of the 10 people sampled prefer saving over spending.

(a) first fill in the details of the binomial probability formula:

p(7)=left(\begin{array}{l}square \\ squareend{array}
ight)(square)^{square}(square)^{square}

(b) the probability, accurate to 4 decimal places, is

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Explanation:

Part (a)

Step1: Identify values for n, k, p

We have \( n = 10 \) (number of trials), \( k = 7 \) (number of successes), and \( p = 0.69 \) (probability of success). Then \( 1 - p = 1 - 0.69 = 0.31 \).

Step2: Substitute into binomial formula

The binomial probability formula is \( P(k)=\binom{n}{k}p^{k}(1 - p)^{n - k} \). Substituting the values, we get \( P(7)=\binom{10}{7}(0.69)^{7}(0.31)^{10 - 7} \), which simplifies to \( P(7)=\binom{10}{7}(0.69)^{7}(0.31)^{3} \).

Step1: Calculate the binomial coefficient

First, calculate \( \binom{10}{7}=\frac{10!}{7!(10 - 7)!}=\frac{10!}{7!3!}=\frac{10\times9\times8}{3\times2\times1}=120 \).

Step2: Calculate \( (0.69)^{7} \) and \( (0.31)^{3} \)

Calculate \( (0.69)^{7}\approx0.69\times0.69\times0.69\times0.69\times0.69\times0.69\times0.69\approx0.0749 \) (more accurately, using a calculator: \( 0.69^7\approx0.074900369 \)).
Calculate \( (0.31)^{3}=0.31\times0.31\times0.31 = 0.029791 \).

Step3: Multiply all components

Multiply the binomial coefficient, \( (0.69)^{7} \), and \( (0.31)^{3} \):
\( 120\times0.074900369\times0.029791\approx120\times0.002231\approx0.2677 \) (after more precise calculation: \( 120\times0.074900369 = 8.98804428 \); \( 8.98804428\times0.029791\approx0.2677 \)).

Answer:

\( P(7)=\boldsymbol{\binom{10}{7}(0.69)^{7}(0.31)^{3}} \) (So the blanks are filled as: \( \binom{10}{7} \), \( 0.69 \), \( 7 \), \( 0.31 \), \( 3 \))

Part (b)