In Families With Four Children You're Interested In The Prob

In Families With Four Children Youre Interested In The Probabilities

In families with four children, you’re interested in the probabilities for the different possible numbers of girls in a family. Using theoretical probability (assume girls and boys are equally likely), compile a five-column table with the headings “0” through “4,” for the five possible numbers of girl children in a four-child family. Then, using “G” for girls and “B” for boys, list under each heading the various birth-order ways of achieving that number of girls in a family. Then, use your table to calculate the following probabilities:

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The probability of having exactly one girl in a family with four children can be calculated by first enumerating all the possible birth orders where only one child is a girl. These arrangements include BBBG, BBGB, BGBB, GBBB, totaling four arrangements. Since each child has an equal chance of being a girl or boy, the total number of combinations for four children is 2^4 = 16. Therefore, the probability of exactly one girl is 4/16 = 1/4 = 25%.

Similarly, the arrangements for exactly two girls include GGBB, GBGB, GBGB, BGBG, totaling six arrangements. The probability is thus 6/16 = 3/8 = 37.5%. For four girls, the only arrangement is GGGG, which has a probability of 1/16 = 6.25%. For the probability that the third child is a girl, note that this event depends on the third child's gender, regardless of other children. Since each child's gender is independent and equally likely, the probability that the third child is a girl is 1/2 or 50%.

Meanwhile, considering a roulette wheel with 38 numbers: 18 odd black, 18 even red, and 2 green (0 and 00), the probabilities of spinning different outcomes are calculated based on the total number of outcomes. The probability of spinning a green number (0 or 00) is 2/38 ≈ 5.26%. The probability of spinning a number greater than 30 (which includes 31 to 36, plus 37 and 38 if considered) depends on the total outcomes greater than 30. Since numbers 31-36, 37, and 38 are 8 out of 38, this probability is 8/38 ≈ 21.05%. The probability of spinning a red number less than 10 includes 2, 4, 6, and 8, totaling 4 outcomes, hence 4/38 ≈ 10.52%. For an even black number, the probability depends on the count of black even numbers; assuming odd black are 1, 3, 5, etc., and similar for red, the probability of an even black number is 0/38 if there are no even black numbers specified, which leads to zero probability if none exist in the set.

The expected total number of green outcomes in 57,000 spins is calculated by multiplying the probability of green by the total spins: 2/38 ≈ 1/19, so E(X) = 57,000 × 1/19 ≈ 3,000 green numbers.

In a survey of 1,500 U.S. residents, 77% said they liked pizza, and one month later, 75% responded positively. To analyze whether pizza's popularity declined, a hypothesis test is performed. The null hypothesis (H0) states that the proportion liking pizza did not decline, so P1 = P2, while the alternative hypothesis (H1) claims that the proportion decreased, P1 > P2. Using sample data, a z-test is computed: with sample sizes of 1500 and sample proportions of 0.77 and 0.75, respectively, the z-statistic is approximately 1.2825, and the corresponding p-value is about 0.0998. Since this p-value exceeds the significance level of 0.05, we fail to reject the null hypothesis, indicating insufficient evidence to support a decline in pizza popularity. However, given the small decline in observed proportions, the result suggests that the difference might not be statistically significant.

In a game involving dice, eleven participants select their favorite number between 2 and 12. Each person bets on their favorite number in 10,000 rolls, with a payoff of $12 if their number comes up, plus an extra dollar from a donor each round, totaling $13 per win. The expected value calculations for the number 7 (which has six combinations out of 36 possible rolls) involve computing the probability: P(7) = 6/36 = 1/6. The expected value (E(X)) for betting on 7 is then 10,000 × [(12 × 1/6) – (1 × 5/6)] = approximately $11,666.66. For the number 2, with probability 1/36, the expected value is 10,000 × [(12 × 1/36) – (1 × 35/36)] ≈ $6,388.88. These calculations demonstrate the expected return based on the probability of each number appearing.

Regarding statistical distributions: when the population is normally distributed and the population standard deviation is known, the z-distribution is used. If the population standard deviation is unknown and the sample size is less than 30, the t-distribution is appropriate. For larger samples with normally distributed populations, the t-distribution is still used if the standard deviation is unknown, and the z-distribution is used if the population standard deviation is known. If the population distribution is not normal and the sample size is small (less than 30), the distribution choice depends on whether the sample size is large enough for the Central Limit Theorem to apply or whether a non-parametric approach is needed.

In a case where a sample of 23 used books at a college bookstore has a mean price of $27.50 and a standard deviation of $6.75, the 95% confidence interval is constructed using the t-distribution. The degrees of freedom are 22 (n-1). The critical t-value for 95% confidence with 22 degrees of freedom is approximately 2.074. The margin of error is calculated as t* × (s/√n) = 2.074 × (6.75/√23) ≈ 2.76. The confidence interval then spans from 27.50 – 2.76 to 27.50 + 2.76, approximately $24.74 to $30.50, indicating that we are 95% confident that the true mean price of used books falls within this range.

From a survey of 354 students at a college, with 44 playing chess (25 males and 19 females), the data aims to compare proportions of male and female students who play chess. The null hypothesis (H0) states that the proportion of male students who play chess equals that of female students, while the alternative hypothesis (H1) suggests a difference exists. This comparison can be extended using techniques like chi-square tests for independence to examine association between gender and chess playing.

In an ANOVA analysis involving three samples, the goal is to determine whether there are statistically significant differences among the means of the three populations. The null hypothesis (H0) claims all three population means are equal, while the alternative hypothesis (H1) states at least one mean differs. The analysis requires sample statistics such as sample means, standard deviations, and sizes. An F-test statistic lower than the critical value indicates failure to reject H0, suggesting no significant difference among the group means.

References

  • Wasserman, L. (2004). All of Statistics: A Concise Course in Statistical Inference. Springer.
  • Siegel, S., & Castellan, N. J. (1988). Nonparametric Statistics for the Behavioral Sciences. McGraw-Hill.
  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics. W.H. Freeman.
  • Devore, J. L. (2011). Probability and Statistics for Engineering and the Sciences. Cengage Learning.
  • Lehmann, E. L., & Romano, J. P. (2005). Testing Statistical Hypotheses. Springer Science & Business Media.
  • Newbold, P., Carlson, W. L., & Thorne, B. (2013). Statistics for Business and Economics. Pearson.
  • Snedecor, G. W., & Cochran, W. G. (1989). Statistical Methods. Iowa State University.
  • Chatterjee, S., & Hadi, A. S. (2015). Regression Analysis by Example. Wiley.
  • Ott, R. L., & Longnecker, M. (2010). An Introduction to Statistical Methods and Data Analysis. Cengage Learning.
  • Zimmerman, D. W. (1997). A Note on the Use of the t Distribution Instead of the Z Distribution in Small Samples. The American Statistician, 51(4), 271-273.

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