A jar contains 3 nickels and 2 dimes. You reach in and randomly select 2 coins. Let x represent the value of the two coins in cents. Find the expected value and standard deviation of x
Let
N=event of picking a nickel (5 cents)
D=event of picking a dime (10 cents)
X()=value of outcome
P()=probability of outcome
Possible outcomes:
NN : X(NN) = 10 P(NN)=(3/5)(2/4)=3/10
ND : X(ND) = 15 P(ND)=(3/5)(2/4)=3/10
DN : X(DN) = 15 P(DN)=(2/5)(3/4)=3/10
DD : X(DD) = 20 P(DD)=(2/5)(1/4)=1/10
E[X]=Σ X(k) P(k) where k=all possible outcomes.
E[X²]=Σ X(k)^2 P(k)
σ²=E[X²]-(E[X])²
Standard deviation = √(σ²)
i have no clue
To find the expected value of x, we need to find the average value of all possible outcomes.
Possible outcomes:
1. Nickel, Nickel: 5 cents + 5 cents = 10 cents
2. Nickel, Dime: 5 cents + 10 cents = 15 cents
3. Dime, Nickel: 10 cents + 5 cents = 15 cents
4. Dime, Dime: 10 cents + 10 cents = 20 cents
Probability of each outcome:
1. Probability of selecting a nickel = Number of nickels / Total number of coins = 3/5
Probability of selecting another nickel = Number of remaining nickels / Remaining coins = 2/4
Probability of outcome (Nickel, Nickel) = (3/5) * (2/4) = 3/10
2. Probability of selecting a nickel = Number of nickels / Total number of coins = 3/5
Probability of selecting a dime = Number of dimes / Remaining coins = 2/4
Probability of outcome (Nickel, Dime) = (3/5) * (2/4) = 3/10
3. Probability of selecting a dime = Number of dimes / Total number of coins = 2/5
Probability of selecting a nickel = Number of remaining nickels / Remaining coins = 3/4
Probability of outcome (Dime, Nickel) = (2/5) * (3/4) = 6/20 = 3/10
4. Probability of selecting a dime = Number of dimes / Total number of coins = 2/5
Probability of selecting another dime = Number of remaining dimes / Remaining coins = 1/4
Probability of outcome (Dime, Dime) = (2/5) * (1/4) = 2/20 = 1/10
Expected value of x = (Value of outcome 1) * (Probability of outcome 1) + (Value of outcome 2) * (Probability of outcome 2) + (Value of outcome 3) * (Probability of outcome 3) + (Value of outcome 4) * (Probability of outcome 4)
Expected value of x = (10 cents * 3/10) + (15 cents * 3/10) + (15 cents * 3/10) + (20 cents * 1/10) = 1 + 4.5 + 4.5 + 2 = 12 cents
To find the standard deviation of x, we need to calculate the variance and then take the square root.
Variance of x = (Value of outcome 1 - Expected value of x)^2 * (Probability of outcome 1)
+ (Value of outcome 2 - Expected value of x)^2 * (Probability of outcome 2)
+ (Value of outcome 3 - Expected value of x)^2 * (Probability of outcome 3)
+ (Value of outcome 4 - Expected value of x)^2 * (Probability of outcome 4)
Variance of x = (10 - 12)^2 * 3/10 + (15 - 12)^2 * 3/10 + (15 - 12)^2 * 3/10 + (20 - 12)^2 * 1/10
= 4 * 3/10 + 9 * 3/10 + 9 * 3/10 + 64 * 1/10
= 1.2 + 2.7 + 2.7 + 6.4
= 13
Standard deviation of x = Square root of variance of x = Square root of 13 ≈ 3.61 cents
To find the expected value and standard deviation of x, we need to determine all the possible outcomes and their respective probabilities.
First, let's list all the possible combinations of selecting two coins from the jar:
1. Selecting two nickels (N, N)
2. Selecting one nickel and one dime (N, D) or (D, N)
3. Selecting two dimes (D, D)
Now, let's assign a value to each outcome:
1. Two nickels: Each nickel is worth 5 cents, so the total value is 5 + 5 = 10 cents.
2. One nickel and one dime: The nickel is worth 5 cents, and the dime is worth 10 cents, so the total value is 5 + 10 = 15 cents.
3. Two dimes: Each dime is worth 10 cents, so the total value is 10 + 10 = 20 cents.
Next, we need to determine the probabilities of each outcome. Since we are selecting two coins without replacement, the probability of each outcome depends on the total number of ways we can select two coins from a jar of 5 coins (3 nickels and 2 dimes).
The total number of ways to select two coins from 5 is given by the combination formula:
C(5, 2) = 5! / (2!(5-2)!) = (5 * 4) / (2 * 1) = 10.
Now, let's determine the probabilities:
1. Probability of selecting two nickels: There are 3 nickels in a jar of 5 coins, so the probability of selecting a nickel on the first draw is 3/5. After selecting a nickel, there are 2 nickels left out of 4 coins, so the probability of selecting another nickel on the second draw is 2/4 (since we have already selected one nickel). The combined probability is (3/5) * (2/4) = 3/10.
2. Probability of selecting one nickel and one dime: Similar to the first case, there are 3 nickels and 2 dimes. The probability of selecting a nickel on the first draw is 3/5, and the probability of selecting a dime on the second draw is 2/4. However, since we can also select a dime on the first draw and a nickel on the second draw, we need to consider both possibilities. Therefore, the combined probability is (3/5) * (2/4) + (2/5) * (3/4) = 6/20 + 6/20 = 12/20 = 3/5.
3. Probability of selecting two dimes: Similarly, the probability of selecting a dime on the first draw is 2/5, and the probability of selecting another dime on the second draw is 1/4. Therefore, the combined probability is (2/5) * (1/4) = 2/20 = 1/10.
Now, let's calculate the expected value (E(x)):
E(x) = (Value of outcome 1) * (Probability of outcome 1) + (Value of outcome 2) * (Probability of outcome 2) + (Value of outcome 3) * (Probability of outcome 3)
E(x) = (10 cents) * (3/10) + (15 cents) * (3/5) + (20 cents) * (1/10)
= 1 + 3 + 2
= 6 cents.
So, the expected value of x is 6 cents.
To calculate the standard deviation, we need to calculate the variance first:
Variance = (Value of outcome 1)^2 * (Probability of outcome 1) + (Value of outcome 2)^2 * (Probability of outcome 2) + (Value of outcome 3)^2 * (Probability of outcome 3) - (E(x))^2
Variance = (10 cents)^2 * (3/10) + (15 cents)^2 * (3/5) + (20 cents)^2 * (1/10) - (6 cents)^2
= 1 + 27 + 8 - 36
= 0 cents.
Since the variance is zero, the standard deviation is also zero.
Therefore, the expected value of x is 6 cents and the standard deviation is 0 cents.