Statistics
Real statistics questions asked by students, solved step by step Every question below was submitted by a real student and answered step by step.
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.
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.
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 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.
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.
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.
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.
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.
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.
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.
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.
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 six cases into one: 1 - C(51,3)/C(60,3), where C(51,3) counts the all-unmarked draws.
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 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.
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 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.