Question

according to a recent survey, 80% of high school students have their own cell phone. suppose you select 10 high school students at random. determine each probability. round your answers to the nearest tenth of a percent if necessary. sample problem p(8 of the students have cell phones) = _{10}c_8 cdot left( \frac{8}{10}
ight)^8 left( \frac{2}{10}
ight)^2 approx 45(0.0067) approx 0.3015 p(5 of the students have cell phones) enter the answer in the space provided. use numbers instead of words. \boxed{}%

Explanation:

Step1: Identify the formula

We use the binomial probability formula \( P(X = k) = {n \choose k} \cdot p^k \cdot (1 - p)^{n - k} \), where \( n = 10 \), \( k = 5 \), \( p = 0.8 \) (since 80% = \( \frac{8}{10} \)) and \( 1 - p = 0.2 \) (since \( 1 - \frac{8}{10} = \frac{2}{10} \)).

Step2: Calculate the combination

First, calculate \( {10 \choose 5} \). The formula for combinations is \( {n \choose k} = \frac{n!}{k!(n - k)!} \). So, \( {10 \choose 5} = \frac{10!}{5!(10 - 5)!} = \frac{10!}{5!5!} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = 252 \).

Step3: Calculate \( p^k \) and \( (1 - p)^{n - k} \)

\( p^k = (0.8)^5 = 0.32768 \) and \( (1 - p)^{n - k} = (0.2)^{5} = 0.00032 \).

Step4: Multiply the three parts together

\( P(5) = {10 \choose 5} \cdot (0.8)^5 \cdot (0.2)^5 = 252 \times 0.32768 \times 0.00032 \). First, multiply \( 0.32768 \times 0.00032 = 0.0001048576 \). Then, multiply by 252: \( 252 \times 0.0001048576 \approx 0.0264241152 \).

Step5: Convert to percentage

To convert to a percentage, multiply by 100: \( 0.0264241152 \times 100 \approx 2.64241152\% \), which rounds to \( 2.6\% \) (to the nearest tenth of a percent).

Answer:

2.6