Question

1. exercise if 37% of high school students said that they exercise regularly, find the probability that 5 randomly selected high school students will say that they exercise regularly. would you consider this event likely or unlikely to occur? explain your answer.

Explanation:

Step1: Identify the probability model

This is a binomial probability problem where each student has a probability \( p = 0.37 \) of saying they exercise regularly, and we want the probability that \( k = 5 \) out of \( n = 5 \) students say they exercise regularly. The binomial probability formula is \( P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \( C(n,k)=\frac{n!}{k!(n - k)!} \).

Step2: Calculate the combination \( C(5,5) \)

For \( n = 5 \) and \( k = 5 \), \( C(5,5)=\frac{5!}{5!(5 - 5)!}=\frac{5!}{5!0!}=1 \) (since \( 0!=1 \)).

Step3: Substitute into the binomial formula

We have \( p = 0.37 \), \( 1-p=1 - 0.37 = 0.63 \), \( n = 5 \), \( k = 5 \). So \( P(X = 5)=C(5,5)\times(0.37)^{5}\times(0.63)^{5 - 5} \). Since \( (0.63)^{0}=1 \) and \( C(5,5) = 1 \), this simplifies to \( P(X = 5)=(0.37)^{5} \).

Step4: Compute \( (0.37)^{5} \)

\( 0.37\times0.37 = 0.1369 \); \( 0.1369\times0.37\approx0.050653 \); \( 0.050653\times0.37\approx0.01874161 \); \( 0.01874161\times0.37\approx0.0069343957 \).

Step5: Determine likelihood

A probability less than 0.05 (or 5%) is often considered unlikely. Since \( 0.006934\approx0.69\%<5\% \), the event is unlikely.

Answer:

The probability that 5 randomly selected high school students will say they exercise regularly is approximately \( 0.0069 \) (or \( 0.69\% \)). This event is unlikely to occur because its probability is less than 5% (a common threshold for determining unlikelihood), meaning the chance of all 5 selected students saying they exercise regularly is very small.