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Indian Navy SSR (Sailor) Statistics and Probability — practice questions

Explore 29 free Statistics and Probability questions in Indian Navy SSR (Sailor) Mathematics. Open each question or exercise to read its content, reference answers and available explanation.

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Statistics and Probability questions and exercises with answers

Showing questions 1–29 of 29.

For two events A and B, P(A | B) = 0.6, P(B) = 0.5, P(A) = 0.4. P(B | A) =Read the answer and worked solution →Mean Absolute Deviation (MAD):Read the answer and worked solution →Two independent X, Y with Var(X) = 4, Var(Y) = 9. Var(X + Y) =Read the answer and worked solution →Tap every measure of dispersion your syllabus listsRead the answer and worked solution →Central tendency or dispersion?Read the answer and worked solution →Complete the definition of rangeRead the answer and worked solution →Working out a rangeRead the answer and worked solution →Order the argument: why mean deviation uses absolute valuesRead the answer and worked solution →True or false: why plain deviations failRead the answer and worked solution →Complete the definition of mean deviationRead the answer and worked solution →Match each measure to how it is builtRead the answer and worked solution →True or false: variance and standard deviationRead the answer and worked solution →Squared or absolute?Read the answer and worked solution →Reading the definitions backRead the answer and worked solution →True or false: what this chapter coversRead the answer and worked solution →Assertion and reasonRead the answer and worked solution →Two are true, one is notRead the answer and worked solution →Three of these are wrong. Find the one that holds.Read the answer and worked solution →Order the steps: compute a mean deviation about the meanRead the answer and worked solution →Order the steps: compute a standard deviationRead the answer and worked solution →Flip and pair: symbol with meaningRead the answer and worked solution →Grouped data or ungrouped data?Read the answer and worked solution →Spot the one that does not belongRead the answer and worked solution →Tap every statement true of the rangeRead the answer and worked solution →Complete the reason absolute values appearRead the answer and worked solution →Central value choicesRead the answer and worked solution →Assertion and reason: choosing a measureRead the answer and worked solution →True or false: measures comparedRead the answer and worked solution →Match each idea to what it answersRead the answer and worked solution →

Worked practice questions

For two events A and B, P(A | B) = 0.6, P(B) = 0.5, P(A) = 0.4. P(B | A) =

  1. 0.75
  2. 0.4
  3. 0.5
  4. 0.6 × 0.5 / 0.4 = 0.75 (Bayes)
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Correct answer: D. 0.6 × 0.5 / 0.4 = 0.75 (Bayes)

We have P(B|A) = P(A|B) × P(B)/P(A). That is P(B|A) = 0.6 × 0.5 / 0.4. So P(B|A) = 0.75.

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Mean Absolute Deviation (MAD):

  1. Sum
  2. Mean
  3. SD
  4. (1/n) Σ |x_i - x̄|
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Correct answer: D. (1/n) Σ |x_i - x̄|

MAD = (1/n) Σ |x_i - mean|. Alternative spread measure (uses absolute values). Less sensitive to outliers than SD. Standard SD penalizes large deviations more.

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Two independent X, Y with Var(X) = 4, Var(Y) = 9. Var(X + Y) =

  1. 13 (independence: variances add)
  2. 6.5
  3. 36
  4. sqrt(13)
Show answer and solution

Correct answer: A. 13 (independence: variances add)

For independent X, Y: Var(X + Y) = Var(X) + Var(Y) = 4 + 9 = 13. (For dependent: Var(X+Y) = Var(X) + Var(Y) + 2Cov(X,Y).)

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