To explain: The result. To explain: Whether the sample is a good one or not. Given info: Here, 200 adults say yes for regularly participating in at least one sport. Also, out of 200 adults who say yes, 9% say they participate in hiking.

Question

To explain: The result.  To explain: Whether the sample is a good one or not.  Given info:  Here, 200 adults say yes for regularly participating in at least one sport. Also, out of 200 adults who say yes, 9% say they participate in hiking.

odgovoreh · Accepted Answer

Calculation: Answers may vary. One of the possible answers is given below: Consider the adults said yes for participate in hiking p is 0.11. The adults said no for participate in hiking q is 0.89.(=1−0.11). The requirements to check whether the normal distribution can be used for approximate the distribution of x, the binomial random variable are np≥5 and nq≥5. That is, Condition 1: np≥5 Condition 2: nq≥5 Condition 1: np≥5 Substitute 200 for n and 0.11 for p in the np as follows: np=200×0.11 =22 &amp;gt;5 Thus, the given requirement np(=22)≥5 is satisfied. Condition 2: nq≥5 Substitute 200 for n and 0.89 for q in the nq as follows: nq=200×0.89 =178 &amp;gt;5 Thus, the given requirement nq(=178)≥5 is satisfied. Since the requirements that np≥5 and nq≥5 are satisfied, the normal distribution can be used to approximate the binomial distribution. That is, the probability from a binomial probability distribution can be approximated by using normal distribution with the parameters are, μ=np and σ=npq. The value of the mean and standard deviation is as follows: μ=200×0.11 =22 σ=200×0.11×0.89 =19.58 =4.4249 From the given information, the percent of adults said yes for participating in hiking game is 0.09. That is the number of persons said yes for participating in hiking is 18(=200×0.09). Therefore, the number of persons said no is 182(=200−18). Thus, the binomial random variable x is less than 18. Continuity correction: If the binomial probability represents &quot;fewer than c&quot; then the normal probability is P(x&amp;lt;c−0.5) for any number c. By using continuity correction, the value 0.5 is subtracted from the value of 18. That is, c=18 P(x&amp;lt;18)=P(x&amp;lt;18−0.5) =P(x&amp;lt;17.5) Here, P(x&amp;lt;17.5) represents the area to the left of 17.5. The formula to convert the x value into z score is as follows: z=x−μσ Substitute 17.5 for x, 22 for μ and 4.4249 for σ as follows: z=17.5−224.4249 =−4.54.4249 =−1.02 Use Table 4: Standard normal distribution to find the area to the left of -1.02. From the table 4, the area of -1.2 is 0.1539. That is, P(z&amp;lt;−1.02)=0.1539 Thus, it can be concluded that the sample is good because it is likely that out of 200 people 18 of them responded that they participate in hiking.

To explain: The result. To explain: Whether the sample is a

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