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Statistical Analysis and Probability Evaluation: Comprehensive Problem Set Solutions

A complete set of solved problems in statistical analysis and probability.

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    Statistical Analysis and Probability Evaluation: Comprehensive Problem Set Solutions Name: Lowenbrau Date: 5 /9/2015 Problem Number Solution 1. (a) FALSE (b) FALSE (c) FALSE (d) TRUE (e) FALS E Work: (a) If all the observations in a data set are identical, then its variance is equals to zero. (b) Since A and B are disjoint, the probability of their intersection must be equal to zero, not equal to 0.9 . (c) For a symmetric distribution if P ( X > 2) = 0.3, then P ( X ≤ – 2) = 0.3, not P ( X ≤ – 3) ≤ 0.3 . (d) The critical value for 9 5 9% confidence level is larger than that of 9 0 % confidence level . So 9 5 % confidence interval has larger width than 9 0 % confidence interval. (e) S ince the p - value for this right tailed test, P (T > 1.5) = 1 – P (X < 1.5) = 1 – 0.96 = 0.04 < 0.05, the decision must be reject the null hypothesis is at 0.05 level of significance. .

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    2. The shaded cells in the following table indicate the calculated values. Check out Time (in minutes) Frequency Relative Frequency 1.0 - 1.9 3 0.12 2.0 - 2.9 12 0.48 3.0 - 3.9 5 0.2 0 4.0 - 4.9 3 0.12 5.0 - 5.9 2 0.08 Total 25 1.00 Work: Total frequency = 25  Value of f when relative frequency is 0.2 0 = 25 X 0.2 0 = 5 . 3 + 12 + 5 + 3 + ? = 25 23 + ? = 25 ? = 25 – 23 = 2 Frequency of the cell Relative frequency Total frequency Relative frequency for 1.0 1.9 3 25 0.12 Relative frequency for 2.0 2.9 12 25 0.48 Relative frequency for 3.0 3.9 0.20 Relative frequency for 4.0               4.9 3 25 0.12 Relative frequency for 5.0 5.9 2 25 0.08       3. Answer: 4 0 % Work: Obtain the percentage of the checkout times was at least 3 minutes, 5 3 2 10 40% 25 25    

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    4. Answer: The median must lie in the class 2 .0 – 2 .9. Work: The following table shows the cumulative relative frequency distribution of the given data. Check out Time Frequency Relative Frequency Cumulative Relative Frequency 1.0 - 1.9 3 0.12 0.12 2.0 - 2.9 12 0.48 0.60 3.0 - 3.9 5 0.2 0.80 4.0 - 4.9 3 0.12 0.92 5.0 - 5.9 2 0.08 1.00 Total 25 1.00 - The median of the distribution covers 50% of the observations in the given dataset. Here the cumulative relative frequency corresponding to 2 .0 – 2 .9 is just greater than 50%, so median must lie in that interval. 5. Answer: This distribution has positive skew . Work: Since the mean of the distribution lie in the class 3.0 – 3.9, and it is greater than the median the distribution has positive skew . 6. Answer: The re quired probability = 1/221 . If the card selection is without replacement, the probability that the first card is an ace and the second card is also an ace is, 4 3 1 52 51 221  

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    7. Answer: The required probability = 1/169 . Work: If the card selection is with replacement, the probability that the first card is an ace and the second card is also an ace is, 4 4 1 52 52 169   8 . Answer: (c) Quiz 1 has the less value of the interquartile range. Work: T he interquartile range for Quiz 1 = 8 5 – 40 = 4 5 . T he interquartile range for Quiz 2 = 9 0 – 35 = 55 . 9 . Answer: ( a ) Quiz 2 Work: For the Quiz 2 the percentage of students with grades 90 or over is 25%, since Q3 = 90 . For the Quiz 1 the percentage of students with grades 90 or ove r is less than 25%, since Q3 = 8 5, indicates that 25% of students are with grades 8 5 or over . So subtracting the percentage between grades 85 and 9 0 , the required percentage is less than 25%. 1 0 . Answer: ( b ) Quiz 2 Work: For the Quiz 1 the percentage of students with grades 60 or less is more than 50 %, since Q 2 = 55 . For the Quiz 2 the percentage of students with grades 6 0 or less is also mor e than 5 0 %, since Q 2 = 50 . But the percentage between grades 5 0 and 60 , is slightly more than that of the percentage between grades 55 and 60, so Quiz 2 has a greater percentage.

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