Study Mean And Standard Deviation in AP Statistics with focused flashcards that help you recognize the idea, recall the key rule, and apply it in practice-style prompts.
All flashcards Flashcard 1: Identify the standard deviation of a Poisson random variable with rate lambda \text{lambda} lambda . Answer: Standard deviation is sqrt(lambda) \text{sqrt(lambda)} sqrt(lambda) . Poisson standard deviation is square root of rate parameter.
Flashcard 2: What is the variance of the difference of two independent random variables X X X and Y Y Y ? Answer: V a r ( X − Y ) = V a r ( X ) + V a r ( Y ) Var(X-Y) = Var(X) + Var(Y) Va r ( X − Y ) = Va r ( X ) + Va r ( Y ) . Variances still add even when subtracting variables.
Flashcard 3: What is the formula for the variance of a linear transformation Y = a + b X Y = a + bX Y = a + b X ? Answer: V a r ( Y ) = b 2 V a r ( X ) Var(Y) = b^2 Var(X) Va r ( Y ) = b 2 Va r ( X ) . Variance multiplied by b 2 b^2 b 2 ; constant a a a has no effect.
Flashcard 4: If X X X is a random variable with variance 25, what is V a r ( 2 X + 3 ) Var(2X + 3) Va r ( 2 X + 3 ) ? Answer: V a r ( 2 X + 3 ) = 100 Var(2X + 3) = 100 Va r ( 2 X + 3 ) = 100 . Constant doesn't affect variance: V a r ( 2 X + 3 ) = 4 ( 25 ) Var(2X + 3) = 4(25) Va r ( 2 X + 3 ) = 4 ( 25 ) .
Flashcard 5: Calculate the variance for P ( X = 1 ) = 0.4 P(X=1)=0.4 P ( X = 1 ) = 0.4 , P ( X = 2 ) = 0.6 P(X=2)=0.6 P ( X = 2 ) = 0.6 . Mean is 1.6 1.6 1.6 . Answer: Variance is 0.24 0.24 0.24 . Sum ( x i − 1.6 ) 2 × p i (x_i - 1.6)^2 \times p_i ( x i − 1.6 ) 2 × p i for each value.
Flashcard 6: Find the variance of a random variable X X X with V a r ( X ) = 9 Var(X)=9 Va r ( X ) = 9 after Y = 2 X − 3 Y=2X-3 Y = 2 X − 3 . Answer: V a r ( Y ) = 36 Var(Y) = 36 Va r ( Y ) = 36 . Apply variance rule: V a r ( Y ) = 2 2 × 9 = 36 Var(Y) = 2^2 \times 9 = 36 Va r ( Y ) = 2 2 × 9 = 36 .
Flashcard 7: What is the variance of a Poisson random variable with rate lambda \text{lambda} lambda ? Answer: V a r ( X ) = lambda Var(X) = \text{lambda} Va r ( X ) = lambda . For Poisson distribution, variance equals the rate parameter.
Flashcard 8: What is the variance of X X X if X X X follows a normal distribution with variance sigma 2 \text{sigma}^2 sigma 2 ? Answer: V a r ( X ) = sigma 2 Var(X) = \text{sigma}^2 Va r ( X ) = sigma 2 . Normal distribution variance is σ 2 \sigma^2 σ 2 by definition.
Flashcard 9: What is the mean of the sum of two independent random variables X X X and Y Y Y ? Answer: E ( X + Y ) = E ( X ) + E ( Y ) E(X+Y) = E(X) + E(Y) E ( X + Y ) = E ( X ) + E ( Y ) . Expected values always add for independent variables.
Flashcard 10: What is the mean of a Poisson random variable with rate lambda \text{lambda} lambda ? Answer: E ( X ) = lambda E(X) = \text{lambda} E ( X ) = lambda . For Poisson distribution, mean equals the rate parameter.
Flashcard 11: For X X X uniform from 2 to 8, calculate the variance. Answer: Variance is 3 3 3 . Uniform variance: ( 8 − 2 ) 2 12 = 36 12 = 3 \frac{(8-2)^2}{12} = \frac{36}{12} = 3 12 ( 8 − 2 ) 2 = 12 36 = 3 .
Flashcard 12: Calculate the mean of a random variable X X X with E ( X ) = 5 E(X) = 5 E ( X ) = 5 after transformation Y = 3 X + 2 Y = 3X + 2 Y = 3 X + 2 . Answer: E ( Y ) = 17 E(Y) = 17 E ( Y ) = 17 . Apply linear transformation: E ( Y ) = 3 ( 5 ) + 2 = 17 E(Y) = 3(5) + 2 = 17 E ( Y ) = 3 ( 5 ) + 2 = 17 .
Flashcard 13: State the formula for the variance of a discrete random variable X X X . Answer: V a r ( X ) = sum of [ ( x i − x ˉ ) 2 × p i ] Var(X) = \text{sum of } [(x_i - \bar{x})^2 \times p_i] Va r ( X ) = sum of [( x i − x ˉ ) 2 × p i ] . Each squared deviation from mean times its probability, then sum.
Flashcard 14: Calculate the variance of X X X with V a r ( X ) = 16 Var(X)=16 Va r ( X ) = 16 after Y = − X Y=-X Y = − X . Answer: V a r ( Y ) = 16 Var(Y) = 16 Va r ( Y ) = 16 . Multiplying by − 1 -1 − 1 doesn't change variance: ( − 1 ) 2 = 1 (-1)^2 = 1 ( − 1 ) 2 = 1 .
Flashcard 15: What is the effect on the mean if X X X is shifted by c c c , i.e., Y = X + c Y = X + c Y = X + c ? Answer: Mean increases by c c c , E ( Y ) = E ( X ) + c E(Y) = E(X) + c E ( Y ) = E ( X ) + c . Adding constant shifts mean but doesn't change spread.
Flashcard 16: What is the effect on the variance if X X X is shifted by c c c , i.e., Y = X + c Y = X + c Y = X + c ? Answer: Variance remains unchanged, V a r ( Y ) = V a r ( X ) Var(Y) = Var(X) Va r ( Y ) = Va r ( X ) . Adding constant doesn't affect variability measures.
Flashcard 17: Calculate the variance for P ( X = 1 ) = 0.4 P(X=1)=0.4 P ( X = 1 ) = 0.4 , P ( X = 2 ) = 0.6 P(X=2)=0.6 P ( X = 2 ) = 0.6 . Mean is 1.6 1.6 1.6 . Answer: Variance is 0.24 0.24 0.24 . Sum ( x i − 1.6 ) 2 × p i (x_i - 1.6)^2 \times p_i ( x i − 1.6 ) 2 × p i for each value.
Flashcard 18: How do you calculate the standard deviation from variance? Answer: Take the square root of the variance, SD ( X ) = sqrt ( V a r ( X ) ) \text{SD}(X) = \text{sqrt}(Var(X)) SD ( X ) = sqrt ( Va r ( X )) . Standard deviation is always the positive square root of variance.
Flashcard 19: What is the mean of X X X if X X X follows a normal distribution with mean mu \text{mu} mu ? Answer: E ( X ) = mu E(X) = \text{mu} E ( X ) = mu . Normal distribution mean is the location parameter μ \mu μ .
Flashcard 20: Identify the standard deviation of X X X if X X X follows a normal distribution with variance sigma 2 \text{sigma}^2 sigma 2 . Answer: Standard deviation is sigma \text{sigma} sigma . Standard deviation is square root of variance: σ 2 = σ \sqrt{\sigma^2} = \sigma σ 2 = σ .
Flashcard 21: What is the mean of the difference of two independent random variables X X X and Y Y Y ? Answer: E ( X − Y ) = E ( X ) − E ( Y ) E(X-Y) = E(X) - E(Y) E ( X − Y ) = E ( X ) − E ( Y ) . Expected values subtract when finding difference.
Flashcard 22: Calculate the mean of a random variable X X X with E ( X ) = 5 E(X) = 5 E ( X ) = 5 after transformation Y = 3 X + 2 Y = 3X + 2 Y = 3 X + 2 . Answer: E ( Y ) = 17 E(Y) = 17 E ( Y ) = 17 . Apply linear transformation: E ( Y ) = 3 ( 5 ) + 2 = 17 E(Y) = 3(5) + 2 = 17 E ( Y ) = 3 ( 5 ) + 2 = 17 .
Flashcard 23: Calculate the variance for X X X with V a r ( X ) = 4 Var(X) = 4 Va r ( X ) = 4 after transformation Y = 3 X + 2 Y = 3X + 2 Y = 3 X + 2 . Answer: V a r ( Y ) = 36 Var(Y) = 36 Va r ( Y ) = 36 . Apply variance rule: V a r ( Y ) = 3 2 × 4 = 36 Var(Y) = 3^2 \times 4 = 36 Va r ( Y ) = 3 2 × 4 = 36 .
Flashcard 24: What is the formula for the variance of a linear transformation Y = a + b X Y = a + bX Y = a + b X ? Answer: V a r ( Y ) = b 2 V a r ( X ) Var(Y) = b^2 Var(X) Va r ( Y ) = b 2 Va r ( X ) . Variance multiplied by b 2 b^2 b 2 ; constant a a a has no effect.
Flashcard 25: What is the variance of a Poisson random variable with rate lambda \text{lambda} lambda ? Answer: V a r ( X ) = lambda Var(X) = \text{lambda} Va r ( X ) = lambda . For Poisson distribution, variance equals the rate parameter.
Flashcard 26: Identify the mean of a uniformly distributed random variable X X X over a a a to b b b . Answer: E ( X ) = a + b 2 E(X) = \frac{a+b}{2} E ( X ) = 2 a + b . Uniform mean is the midpoint of the interval.
Flashcard 27: What is the mean of a binomial random variable X X X with n n n trials and probability p p p ? Answer: E ( X ) = n × p E(X) = n \times p E ( X ) = n × p . Binomial mean equals number of trials times success probability.
Flashcard 28: State the formula for the mean of a linear transformation Y = a + b X Y = a + bX Y = a + b X . Answer: E ( Y ) = a + b E ( X ) E(Y) = a + bE(X) E ( Y ) = a + b E ( X ) . Linear transformation: add constant a a a , multiply by b b b .
Flashcard 29: What is the variance of the difference of two independent random variables X X X and Y Y Y ? Answer: V a r ( X − Y ) = V a r ( X ) + V a r ( Y ) Var(X-Y) = Var(X) + Var(Y) Va r ( X − Y ) = Va r ( X ) + Va r ( Y ) . Variances still add even when subtracting variables.
Flashcard 30: What is the variance of X X X if X X X follows a normal distribution with variance sigma 2 \text{sigma}^2 sigma 2 ? Answer: V a r ( X ) = sigma 2 Var(X) = \text{sigma}^2 Va r ( X ) = sigma 2 . Normal distribution variance is σ 2 \sigma^2 σ 2 by definition.
Flashcard 31: What is the variance of X X X if X X X follows an exponential distribution with rate lambda \text{lambda} lambda ? Answer: V a r ( X ) = 1 lambda 2 Var(X) = \frac{1}{\text{lambda}^2} Va r ( X ) = lambda 2 1 . Exponential variance is reciprocal of rate squared.
Flashcard 32: For a random variable X X X , what happens to the standard deviation if X X X is multiplied by b b b ? Answer: Standard deviation is multiplied by ∣ b ∣ |b| ∣ b ∣ . Multiplying by b b b scales standard deviation by ∣ b ∣ |b| ∣ b ∣ .
Flashcard 33: What is the variance of X X X if X X X is a geometric random variable with probability p p p ? Answer: V a r ( X ) = 1 − p p 2 Var(X) = \frac{1-p}{p^2} Va r ( X ) = p 2 1 − p . Geometric variance uses ( 1 − p ) (1-p) ( 1 − p ) in numerator, p 2 p^2 p 2 in denominator.
Flashcard 34: What is the effect on the mean if X X X is shifted by c c c , i.e., Y = X + c Y = X + c Y = X + c ? Answer: Mean increases by c c c , E ( Y ) = E ( X ) + c E(Y) = E(X) + c E ( Y ) = E ( X ) + c . Adding constant shifts mean but doesn't change spread.
Flashcard 35: State the formula for the variance of a discrete random variable X X X . Answer: V a r ( X ) = sum of [ ( x i − x ˉ ) 2 × p i ] Var(X) = \text{sum of } [(x_i - \bar{x})^2 \times p_i] Va r ( X ) = sum of [( x i − x ˉ ) 2 × p i ] . Each squared deviation from mean times its probability, then sum.
Flashcard 36: Calculate the variance of X X X with V a r ( X ) = 16 Var(X)=16 Va r ( X ) = 16 after Y = − X Y=-X Y = − X . Answer: V a r ( Y ) = 16 Var(Y) = 16 Va r ( Y ) = 16 . Multiplying by − 1 -1 − 1 doesn't change variance: ( − 1 ) 2 = 1 (-1)^2 = 1 ( − 1 ) 2 = 1 .
Flashcard 37: For a random variable X X X , what happens to the standard deviation if X X X is multiplied by b b b ? Answer: Standard deviation is multiplied by ∣ b ∣ |b| ∣ b ∣ . Multiplying by b b b scales standard deviation by ∣ b ∣ |b| ∣ b ∣ .
Flashcard 38: Calculate the mean of X X X if X X X is uniform from 1 to 5. Answer: E ( X ) = 3 E(X) = 3 E ( X ) = 3 . Uniform from 1 to 5: mean = ( 1 + 5 ) / 2 = 3 (1+5)/2 = 3 ( 1 + 5 ) /2 = 3 .
Flashcard 39: Find the mean of a random variable X X X with E ( X ) = 4 E(X)=4 E ( X ) = 4 after Y = 2 X − 3 Y=2X-3 Y = 2 X − 3 . Answer: E ( Y ) = 5 E(Y) = 5 E ( Y ) = 5 . Apply transformation: E ( Y ) = 2 ( 4 ) − 3 = 5 E(Y) = 2(4) - 3 = 5 E ( Y ) = 2 ( 4 ) − 3 = 5 .
Flashcard 40: Identify the mean of a uniformly distributed random variable X X X over a a a to b b b . Answer: E ( X ) = a + b 2 E(X) = \frac{a+b}{2} E ( X ) = 2 a + b . Uniform mean is the midpoint of the interval.
Flashcard 41: For X X X uniform from 2 to 8, calculate the variance. Answer: Variance is 3 3 3 . Uniform variance: ( 8 − 2 ) 2 12 = 36 12 = 3 \frac{(8-2)^2}{12} = \frac{36}{12} = 3 12 ( 8 − 2 ) 2 = 12 36 = 3 .
Flashcard 42: Find the mean of a random variable X X X given P ( X = 1 ) = 0.2 P(X=1)=0.2 P ( X = 1 ) = 0.2 , P ( X = 2 ) = 0.5 P(X=2)=0.5 P ( X = 2 ) = 0.5 , P ( X = 3 ) = 0.3 P(X=3)=0.3 P ( X = 3 ) = 0.3 . Answer: E ( X ) = 2.1 E(X) = 2.1 E ( X ) = 2.1 . Sum each value times its probability: 1 ( 0.2 ) + 2 ( 0.5 ) + 3 ( 0.3 ) 1(0.2) + 2(0.5) + 3(0.3) 1 ( 0.2 ) + 2 ( 0.5 ) + 3 ( 0.3 ) .
Flashcard 43: What is the formula for the expected value of a discrete random variable X X X ? Answer: E ( X ) = sum of ( x i × p i ) E(X) = \text{sum of } (x_i \times p_i) E ( X ) = sum of ( x i × p i ) . Each value times its probability, then sum all products.
Flashcard 44: Find the mean of a random variable X X X given P ( X = 1 ) = 0.2 P(X=1)=0.2 P ( X = 1 ) = 0.2 , P ( X = 2 ) = 0.5 P(X=2)=0.5 P ( X = 2 ) = 0.5 , P ( X = 3 ) = 0.3 P(X=3)=0.3 P ( X = 3 ) = 0.3 . Answer: E ( X ) = 2.1 E(X) = 2.1 E ( X ) = 2.1 . Sum each value times its probability: 1 ( 0.2 ) + 2 ( 0.5 ) + 3 ( 0.3 ) 1(0.2) + 2(0.5) + 3(0.3) 1 ( 0.2 ) + 2 ( 0.5 ) + 3 ( 0.3 ) .
Flashcard 45: State the formula for the mean of a linear transformation Y = a + b X Y = a + bX Y = a + b X . Answer: E ( Y ) = a + b E ( X ) E(Y) = a + bE(X) E ( Y ) = a + b E ( X ) . Linear transformation: add constant a a a , multiply by b b b .
Flashcard 46: What is the variance of a uniformly distributed random variable X X X over a a a to b b b ? Answer: V a r ( X ) = ( b − a ) 2 12 Var(X) = \frac{(b-a)^2}{12} Va r ( X ) = 12 ( b − a ) 2 . Uniform variance uses range squared divided by 12.
Flashcard 47: What is the standard deviation of a binomial random variable X X X with n n n trials and probability p p p ? Answer: SD ( X ) = n × p × ( 1 − p ) \text{SD}(X) = \sqrt{n \times p \times (1-p)} SD ( X ) = n × p × ( 1 − p ) . Binomial standard deviation uses n n n , p p p , and ( 1 − p ) (1-p) ( 1 − p ) under square root.
Flashcard 48: How do you calculate the standard deviation from variance? Answer: Take the square root of the variance, SD ( X ) = sqrt ( V a r ( X ) ) \text{SD}(X) = \text{sqrt}(Var(X)) SD ( X ) = sqrt ( Va r ( X )) . Standard deviation is always the positive square root of variance.
Flashcard 49: Find the standard deviation for V a r ( X ) = 4 Var(X)=4 Va r ( X ) = 4 . Answer: SD ( X ) = 2 \text{SD}(X) = 2 SD ( X ) = 2 . Standard deviation is the square root of variance.
Flashcard 50: What is the mean of X X X if X X X is a geometric random variable with probability p p p ? Answer: E ( X ) = 1 p E(X) = \frac{1}{p} E ( X ) = p 1 . Geometric mean is reciprocal of success probability.
Flashcard 51: Calculate the mean of X X X if X X X is uniform from 1 to 5. Answer: E ( X ) = 3 E(X) = 3 E ( X ) = 3 . Uniform from 1 to 5: mean = ( 1 + 5 ) / 2 = 3 (1+5)/2 = 3 ( 1 + 5 ) /2 = 3 .
Flashcard 52: Calculate the mean of X X X with E ( X ) = 3 E(X)=3 E ( X ) = 3 after Y = − X Y=-X Y = − X . Answer: E ( Y ) = − 3 E(Y) = -3 E ( Y ) = − 3 . Multiplying by − 1 -1 − 1 changes sign: E ( − X ) = − E ( X ) E(-X) = -E(X) E ( − X ) = − E ( X ) .
Flashcard 53: What is the mean of the difference of two independent random variables X X X and Y Y Y ? Answer: E ( X − Y ) = E ( X ) − E ( Y ) E(X-Y) = E(X) - E(Y) E ( X − Y ) = E ( X ) − E ( Y ) . Expected values subtract when finding difference.
Flashcard 54: What is the mean of X X X if X X X is a geometric random variable with probability p p p ? Answer: E ( X ) = 1 p E(X) = \frac{1}{p} E ( X ) = p 1 . Geometric mean is reciprocal of success probability.
Flashcard 55: What is the mean of X X X if X X X follows a normal distribution with mean mu \text{mu} mu ? Answer: E ( X ) = mu E(X) = \text{mu} E ( X ) = mu . Normal distribution mean is the location parameter μ \mu μ .
Flashcard 56: What is the standard deviation of a binomial random variable X X X with n n n trials and probability p p p ? Answer: SD ( X ) = sqrt ( n × p × ( 1 − p ) ) \text{SD}(X) = \text{sqrt}(n \times p \times (1-p)) SD ( X ) = sqrt ( n × p × ( 1 − p )) . Binomial standard deviation uses n n n , p p p , and ( 1 − p ) (1-p) ( 1 − p ) under square root.
Flashcard 57: What is the variance of X X X if X X X is a geometric random variable with probability p p p ? Answer: V a r ( X ) = 1 − p p 2 Var(X) = \frac{1-p}{p^2} Va r ( X ) = p 2 1 − p . Geometric variance uses ( 1 − p ) (1-p) ( 1 − p ) in numerator, p 2 p^2 p 2 in denominator.
Flashcard 58: Calculate the variance for X X X with V a r ( X ) = 4 Var(X) = 4 Va r ( X ) = 4 after transformation Y = 3 X + 2 Y = 3X + 2 Y = 3 X + 2 . Answer: V a r ( Y ) = 36 Var(Y) = 36 Va r ( Y ) = 36 . Apply variance rule: V a r ( Y ) = 3 2 × 4 = 36 Var(Y) = 3^2 \times 4 = 36 Va r ( Y ) = 3 2 × 4 = 36 .
Flashcard 59: Calculate the mean of X X X with E ( X ) = 3 E(X)=3 E ( X ) = 3 after Y = − X Y=-X Y = − X . Answer: E ( Y ) = − 3 E(Y) = -3 E ( Y ) = − 3 . Multiplying by − 1 -1 − 1 changes sign: E ( − X ) = − E ( X ) E(-X) = -E(X) E ( − X ) = − E ( X ) .
Flashcard 60: If X X X is a random variable with variance 25, what is V a r ( 2 X + 3 ) Var(2X + 3) Va r ( 2 X + 3 ) ? Answer: V a r ( 2 X + 3 ) = 100 Var(2X + 3) = 100 Va r ( 2 X + 3 ) = 100 . Constant doesn't affect variance: V a r ( 2 X + 3 ) = 4 ( 25 ) Var(2X + 3) = 4(25) Va r ( 2 X + 3 ) = 4 ( 25 ) .
Flashcard 61: Find the mean of a random variable X X X with E ( X ) = 4 E(X)=4 E ( X ) = 4 after Y = 2 X − 3 Y=2X-3 Y = 2 X − 3 . Answer: E ( Y ) = 5 E(Y) = 5 E ( Y ) = 5 . Apply transformation: E ( Y ) = 2 ( 4 ) − 3 = 5 E(Y) = 2(4) - 3 = 5 E ( Y ) = 2 ( 4 ) − 3 = 5 .
Flashcard 62: Find the variance of a random variable X X X with V a r ( X ) = 9 Var(X)=9 Va r ( X ) = 9 after Y = 2 X − 3 Y=2X-3 Y = 2 X − 3 . Answer: V a r ( Y ) = 36 Var(Y) = 36 Va r ( Y ) = 36 . Apply variance rule: V a r ( Y ) = 2 2 × 9 = 36 Var(Y) = 2^2 \times 9 = 36 Va r ( Y ) = 2 2 × 9 = 36 .
Flashcard 63: Identify the standard deviation of a Poisson random variable with rate λ \lambda λ . Answer: Standard deviation is λ \sqrt{\lambda} λ . Poisson standard deviation is square root of rate parameter.
Flashcard 64: What is the mean of a Poisson random variable with rate lambda \text{lambda} lambda ? Answer: E ( X ) = lambda E(X) = \text{lambda} E ( X ) = lambda . For Poisson distribution, mean equals the rate parameter.
Flashcard 65: What is the mean of X X X if X X X follows an exponential distribution with rate lambda \text{lambda} lambda ? Answer: E ( X ) = 1 lambda E(X) = \frac{1}{\text{lambda}} E ( X ) = lambda 1 . Exponential mean is reciprocal of rate parameter.
Flashcard 66: What is the variance of X X X if X X X follows an exponential distribution with rate lambda \text{lambda} lambda ? Answer: V a r ( X ) = 1 lambda 2 Var(X) = \frac{1}{\text{lambda}^2} Va r ( X ) = lambda 2 1 . Exponential variance is reciprocal of rate squared.
Flashcard 67: What is the variance of the sum of two independent random variables X X X and Y Y Y ? Answer: V a r ( X + Y ) = V a r ( X ) + V a r ( Y ) Var(X+Y) = Var(X) + Var(Y) Va r ( X + Y ) = Va r ( X ) + Va r ( Y ) . Variances add for independent variables (never subtract).
Flashcard 68: If X X X is a random variable with mean 10, what is E ( 2 X + 3 ) E(2X + 3) E ( 2 X + 3 ) ? Answer: E ( 2 X + 3 ) = 23 E(2X + 3) = 23 E ( 2 X + 3 ) = 23 . Apply linear transformation: E ( 2 X + 3 ) = 2 ( 10 ) + 3 E(2X + 3) = 2(10) + 3 E ( 2 X + 3 ) = 2 ( 10 ) + 3 .
Flashcard 69: If X X X is a random variable with mean 10, what is E ( 2 X + 3 ) E(2X + 3) E ( 2 X + 3 ) ? Answer: E ( 2 X + 3 ) = 23 E(2X + 3) = 23 E ( 2 X + 3 ) = 23 . Apply linear transformation: E ( 2 X + 3 ) = 2 ( 10 ) + 3 E(2X + 3) = 2(10) + 3 E ( 2 X + 3 ) = 2 ( 10 ) + 3 .
Flashcard 70: What is the variance of a uniformly distributed random variable X X X over a a a to b b b ? Answer: V a r ( X ) = ( b − a ) 2 12 Var(X) = \frac{(b-a)^2}{12} Va r ( X ) = 12 ( b − a ) 2 . Uniform variance uses range squared divided by 12.
Flashcard 71: What is the variance of the sum of two independent random variables X X X and Y Y Y ? Answer: V a r ( X + Y ) = V a r ( X ) + V a r ( Y ) Var(X+Y) = Var(X) + Var(Y) Va r ( X + Y ) = Va r ( X ) + Va r ( Y ) . Variances add for independent variables (never subtract).
Flashcard 72: Identify the standard deviation of X X X if X X X follows a normal distribution with variance sigma 2 \text{sigma}^2 sigma 2 . Answer: Standard deviation is sigma \text{sigma} sigma . Standard deviation is square root of variance: σ 2 = σ \sqrt{\sigma^2} = \sigma σ 2 = σ .
Flashcard 73: What is the formula for the expected value of a discrete random variable X X X ? Answer: E ( X ) = sum of ( x i × p i ) E(X) = \text{sum of } (x_i \times p_i) E ( X ) = sum of ( x i × p i ) . Each value times its probability, then sum all products.
Flashcard 74: What is the mean of a binomial random variable X X X with n n n trials and probability p p p ? Answer: E ( X ) = n × p E(X) = n \times p E ( X ) = n × p . Binomial mean equals number of trials times success probability.
Flashcard 75: What is the mean of the sum of two independent random variables X X X and Y Y Y ? Answer: E ( X + Y ) = E ( X ) + E ( Y ) E(X+Y) = E(X) + E(Y) E ( X + Y ) = E ( X ) + E ( Y ) . Expected values always add for independent variables.
Flashcard 76: Find the standard deviation for V a r ( X ) = 4 Var(X)=4 Va r ( X ) = 4 . Answer: SD ( X ) = 2 \text{SD}(X) = 2 SD ( X ) = 2 . Standard deviation is the square root of variance.
Flashcard 77: What is the mean of X X X if X X X follows an exponential distribution with rate lambda \text{lambda} lambda ? Answer: E ( X ) = 1 lambda E(X) = \frac{1}{\text{lambda}} E ( X ) = lambda 1 . Exponential mean is reciprocal of rate parameter.