Statistics
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Real statistics questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
With 5% of men and 0.25% of women colour blind and equal numbers of each, total probability is 0.02625, and Bayes' theorem gives P(male | colour blind) = 0.025/0.02625 = 20/21 ≈ 0.952.
An email containing the trigger word is spam with probability 7/12 ≈ 58.3%. Work through the prior, the two likelihoods and Bayes' rule step by step.
The answers are 0.219, 0.145 and 0.855. With n = 8 and p = 0.5 every outcome has weight 1/256, so the counts 56, 37 and 219 give the three probabilities.
P(X < 4) is about 0.022 when n = 10 and p = 0.66. Sum the four terms for X = 0, 1, 2 and 3; each uses a binomial coefficient times 0.66^k times 0.34^(10-k).
P(5) = 0.219, P(X > 5) = 0.145 and P(X <= 5) = 0.855. With p = 0.5 every outcome is equally likely, so each answer is a count from Pascal's triangle over 256.
Chi-square is exactly 5.84 on 2 degrees of freedom, below the 5.991 critical value, so the predicted 3:2:1 ratio is not rejected at the 0.05 significance level.
Means 2.43 and 1.86 differ by 0.57. Pooling SS = 17.71 and 10.86 over 12 degrees of freedom gives s = 1.543, so d = 0.57/1.543 = 0.37.
The effect size is d = 1.33, a large effect. Get SS = 351 - 48^2/8 = 63, then s = 3, and divide the 4-point mean difference by s - not by the standard error.
Both happening gives 0.73^2 = 53.29%; at least one gives 1 - 0.27^2 = 92.71%. The word combined is ambiguous, and the two readings differ by 39 points.
Set B has sigma 3.29 versus set A at 2.35, so B is more variable. Identical means and sizes make the sum of squared deviations, 184 against 94, the whole story.
The probability is 0.456. Conditioning on "Black" restricts the sample space to that row's total of 250, so divide the 114 two-adult-generation families by 250, not by 1438.
The probability is 0.535. Conditioning on "skip generation" makes the column total 99 the denominator, and the White-and-skip cell contributes 53, giving 53/99 = 0.535.
Income is explanatory and happiness is the response. The row percentages 12.8/61.4/25.8, 13.3/64.0/22.7 and 13.2/58.8/28.0 are nearly identical, so the association is weak.
The 95% confidence interval is (3.94, 13.72). Work through the standard error 41.40/sqrt(278) = 2.483, the t critical value 1.969, and the margin of error 4.89.
Means are 70.43, 69.70 and 78.54 with sample SDs 20.67, 20.17 and 13.80. Two students scored 0 on the first two columns, inflating both ranges to 90 and 92.
For the 21 values near 2.4 the mean is 2.4405, median 2.44, range 0.16, and s = 0.0432. The set is trimodal at 2.40, 2.43 and 2.49 — all five measures worked out.
The counts are 63, 15 and 39. Stars and bars handles each item type separately, and the answers multiply because candy and cookie choices are independent.
The seven friends average $2,179.29, higher than the $2,110 estimate, and Ashley would need to spend $1,625 to drag the eight-person mean back down to $2,110.
For the low group the frequencies for scores 1 to 5 are 3, 5, 4, 2, 1; for the high group 0, 1, 3, 6, 5. Each column must total 15 - a check that catches the common 1-versus-3 miscount.
IQR = 39.46, so the fences are -41.17 and 116.67 and the maximum of 248 is an outlier. A mean of 48.00 above the median of 32.50 also marks the data as skewed.
Normalising over the triangle gives A = 2. Splitting the region at x = 1/2 gives P(2X - Y < 1) = 7/12, and integrating over the corner rectangle gives F(1/2, 1/4) = 19/64, while F(1/2, -1) = 0.
Normalising gives c = 3. The marginals are 3x^2 on (0,1] and (3/2)(1 - y^2) on [0,1], and their product is not the joint density, so X and Y are not independent.
Normalising over the region between the parabola y = x^2 and the line y = 1 gives c = 21/4. The marginals are (21/8)x^2(1 - x^4) on [-1,1] and (7/2)y^(5/2) on [0,1].
Averaging the fifteen weights gives 77.81 kg before and 76.20 kg after. The sample standard deviations (dividing by n - 1 = 14) are 15.94 and 16.19 kg.
The mean is 2.23 absences, the variance is 2.3571, and P(more than two absent) = 0.52. Use E(X^2) - mu^2 rather than expanding every squared deviation.
E(X) = 1 and Var(X) = 1/6 for the density f(x) = x on (0,1] and 2 - x on [1,2). Symmetry gives the mean instantly; E(X^2) = 7/6 then delivers the variance.
The mean is 679.63 dollars, the median 689.50 and the mode 699. The eight prices total 5,437, so a printed mean of 670.63 is a transposition error.
Answer: mean 31/6 = 5.1667, median 5, mode 5. Add and divide for the mean, sort first and average the two middle values for the median, then count the repeats.
The mean is 17/3 = 5.67 and the median is 5.5. Every value occurs once, so this data set has no mode - a case worth recognising explicitly.
The mean is 2,179.29 dollars, above the 2,110 estimate. To pull an eight-value average down to 2,110 the extra amount must be 8(2110) - 15,255 = 1,625 dollars.
The mean is greater than the median. The median is fixed at 49, while the mean equals 49 + 22k/N, so the few high scores of 80 pull it strictly upward.
The mean is less than the median. The median is exactly 76 for any class size, while the mean equals 76 − 19k/N and so falls short of it.
The mean is less than the median. The median is pinned at 82.5 while the mean is 82.5 − 19k/N, so the few low scores of 54 drag it below.
The means are $47.33 for the year, $43.00 for January to June and $50.25 for April to November - the summation limits are what change, not the formula.
Play rock 1/4, paper 1/8, scissors 5/8. Equalising the three pure-strategy payoffs gives 5q − r = −5p + 2r = p − 2q = 0, and the value of the symmetric game is 0.
The interval is -0.48 to 4.48. With n = 3 the t critical value is 4.303 and the standard error is 1/sqrt(3), so the margin of error is a wide 2.48.
Test times are normal with mean 58 minutes and standard deviation 12. Find the time by which 90% of students finish, using the z-score for the 90th percentile.
The probability is about 0.3725. The z-score is +0.3253, the left-tail area is 0.62752, and subtracting that from 1 gives the right-tail probability of 0.37248.
The probability is about 0.0785. Standardising gives z = (40.851 − 55)/10 = −1.4149, and the standard-normal area to the left of that z-score is 0.078549.
Going only east or south, every route is 3 east moves and 3 south moves in some order, so the count is the number of ways to place the 3 east moves among 6, which is 20.
The result is t = -2.83. Convert the sums to a mean of 6 and s = 3, divide by the standard error 3/sqrt(8) = 1.0607, and keep the negative sign.
With N = 8, sum X = 48 and sum X^2 = 351, the mean is 6, s = 3 and t = 3.771 on 7 df (p = 0.007). Reject the null at .05 and .01, and Cohen's d = 1.33.
Eta squared is 43.14/86.57 = 0.498, about 0.50. Build SS total and SS between from the group sums of X and X^2, then finish with F(2, 18) = 8.94, p = 0.002.
Answer: p is about 9.96 x 10^-32. Convert the chi-square statistic to a z-score of 11.72, then use the two-tailed normal tail to get the exact tail probability.
Mean BMI fell 0.86 for 15 participants; the paired t-test gives t = 2.363 on 14 df and p = 0.033, so the reduction is significant at the 5% level. Full working.
Sigma is about 2.35, from a sum of squared deviations of 94 over 17 values. Because the mean is exactly 12 every deviation is a whole number, so no rounding creeps in.
The probability is 15796/34220 = 46.16%. By exchangeability the 3 cards drawn after an unseen opening hand behave like 3 cards drawn from the full 60-card deck.
Answer: P(all five occur) = 0.2^5 = 0.00032, about 0.032 percent, and the expected number that occur is np = 5 x 0.2 = 1 by the binomial mean formula.
For independent events, use the complement 1 - (0.8)(0.7)(0.5) = 0.72 for at least one hit, and add the three mutually exclusive single-hit products for exactly one.
The probability is 3/28, about 10.7%. Count the C(3,2) = 3 favourable pairs against the C(8,2) = 28 possible pairs, or multiply 3/8 by 2/7 draw by draw.
Answer: 791/792 = 0.99874. Draw 5 balls from 5 red and 7 blue; subtract the single all-red selection from the C(12,5) = 792 equally likely samples.
The answer is 923/924, about 99.89%. Only one of the 924 possible 6-ball selections misses blue entirely, so the complement rule gives the result in one line.
Drawing 5 cards from 60 with 5 marked A and 23 marked B, the chance of at least one of each is 1748230/5461512 = 32.01%. Worked with inclusion-exclusion.
The answer is 13395/34220, about 39.14%. The complement rule turns three cases into one: 1 - C(51,3)/C(60,3), where C(51,3) counts the all-unmarked draws.
The probability is 1 - (5/6)^8, about 0.7674. 'At least one' is the complement of 'none', and the chance of avoiding a 6 eight times in a row is (5/6)^8.
The probability is 887/924 = 0.96. Counting 'at least 2' directly needs five cases, so use the complement: 0 reds or 1 red account for only 37 of 924 draws.
The probability is 1295/1296, about 0.99923. The complement of 'at most three' is 'all four', a single outcome with probability (1/6)^4 = 1/1296.
The probability is 4/52 = 1/13, about 0.0769 or 7.69%. Count the four kings against 52 equally likely cards, then confirm with the 4:48 odds ratio.
This is binomial with n = 5 and p = 1/3: C(5,1)(1/3)(2/3)^4 = 5 x 16/243 = 80/243, about 0.329.
Use the hypergeometric count C(3,k)C(12,5-k)/C(15,5). Exactly two defectives gives 660/3003 = 20/91, and since only three defectives exist, at least two adds the 66 three-defective draws for 22/91.
The probability is 30/44 = 15/22 ≈ 0.6818, or about 68.18%. Divide favourable outcomes by total outcomes, then reduce the fraction by the common factor 2.
The probability is 1/6 ≈ 0.1667, about 16.67%. A fair die has six equally likely outcomes and exactly one of them is a 6, so the ratio is one favourable case out of six.
b = 3 means at least two of the five draws show a 3, so subtract the zero-3 and one-3 binomial terms from 1: 1 - 32/243 - 80/243 = 131/243.
Fix the number 5 as a member, then choose the other two from above it (C(5,2)=10) or below it (C(4,2)=6), out of C(10,3)=120 equally likely triples.
For (10, 2.9), (20, 4.0), (30, 4.9) the fit is y = -0.001x^2 + 0.14x + 1.6 and R^2 = 1: three points sit exactly on a parabola, so every residual is zero.
Tukey's hinges give Q1 = 27 and Q3 = 39; excluding the median gives 26.5 and 40.5. A worked example on 25 ages showing exactly why the two conventions differ.
The sample standard deviation is about 0.249, from mean 5.7417 and a sum of squares of 0.3099. See why the single high value 6.19 supplies most of the spread.
The seventh quote must be $19,400. Multiply the target mean by 7 to get the required total of $129,500, then subtract the $110,100 already quoted.
The solution is a = −0.0025, b = 0.325, c = 2.9. Subtracting consecutive equations kills c and leaves 30a + b = 0.25 and 50a + b = 0.2, so 20a = −0.05.
The rearranged formula is m = x - zd. Multiply both sides by the standard deviation first, then isolate m; a positive z means the value sits above the mean.
The sample standard deviation is s ≈ 0.2309 and the population value is σ ≈ 0.1886. The mean 2.7667 repeats, so keep extra digits before squaring the deviations.
The population standard deviation is 1.72 and the sample value is 1.92. Mean 5.2, squared deviations summing to 14.8 — see which divisor, n or n − 1, applies.
Population sigma is 0.0125 and the sample s is 0.0132, from mean 4.25 and a sum of squares of just 0.0014. Commas in the source data are decimal points, not separators.
The sample standard deviation is s ≈ 0.005568 and the population value is σ ≈ 0.004546. The mean is exactly 0.651, and the deviations are −0.006, +0.005 and +0.001.
The sample standard deviations are 11.81 and 12.30 percentage points, with means 24.96 and 23.71. The printed before-mean of 24.99 is wrong; the fifteen values actually average 24.96.
The standard error is 1.414, so none of the offered options is correct. SS = 60 - 100/5 = 40 gives s = 3.162, and dividing by sqrt(5) leaves sqrt(2).
The standard error is 1.581. Compute SS = 46 - 8^2/4 = 30, divide by n - 1 = 3 to get s = 3.162, then divide that by sqrt(4) = 2 to reach 1.581.
Tukey's hinge method keeps the median in both halves, so each half has 13 values and the quartiles are their 7th values: 27 and 36 - not the 26 and 39 that the exclusive method gives.
The p-value is about 0.0357. The upper tail beyond z = 2.1 holds 0.0179 of the area, and a two-tailed test doubles it to count the mirror tail below -2.1.
It means 95 percent of intervals built this way capture the true mean. The interval 14 to 17 either contains mu or does not; the level describes the method.
z = (x - mean)/SD gives -2.133 for the youngest customer and 1.580 for the oldest. The asymmetry of the two z-scores is itself evidence of left skew in this sample.