Math  /  Data & Statistics

Questionh Jay I Wh. Untitied document -.. Time Remaining: 53:18 Miaterm Exam: Ch 1(1,2)2(1,2,3)3(1,2,3)4(1,2)5(1,2,3)6(1,2)1(1,2) 2(1,2,3) 3(1,2,3) 4(1,2) 5(1,2,3) 6(1,2) Question 28 of 30 (1 point) I Question Attempts 1 of 1 19 =20=20 21\equiv 21 22\equiv 22 23 =24=24 =26=26 27\equiv 27 28\equiv 28 29
Take a guess: A student takes a multiple-choice test that has 11 questions. Each question has five choices. The student guesses randomly at each answer. Let XX be the number of questions answered correctly. Round the answers to at least four decimal places.
Part: 0/20 / 2
Part 1 of 2 (a) P(6)=0.0097P(6)=0.0097
Part: 1/21 / 2
Part 2 of 2 (b) P(P( More than 3)=)= \square

Studdy Solution

STEP 1

What is this asking? If a student randomly guesses on a multiple-choice test with 11 questions and 5 choices each, what's the chance they get more than 3 questions right? Watch out! Don't forget that "more than 3" doesn't include 3 itself!
Also, each guess is independent, meaning one guess doesn't affect another.

STEP 2

1. Define the distribution
2. Calculate the probability of *not* getting more than 3 right
3. Calculate the probability of getting more than 3 right

STEP 3

We're dealing with a binomial distribution here because we have a fixed number of trials (n=11n = 11 questions), each trial has two outcomes (right or wrong), the probability of success (guessing correctly) is the same for each trial (p=15=0.2p = \frac{1}{5} = 0.2), and the trials are independent.
This is exactly what a binomial distribution is all about!

STEP 4

The probability of getting *not* more than 3 right means getting 0, 1, 2, or 3 right.
We can write this as P(X3)P(X \le 3).
We'll calculate each of these probabilities separately using the binomial probability formula: P(X=k)=(nk)pk(1p)nkP(X=k) = \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k}.
Remember, this formula tells us the probability of getting exactly kk successes in nn trials.

STEP 5

P(X=0)=(110)(0.2)0(0.8)110.0859P(X=0) = \binom{11}{0} \cdot (0.2)^0 \cdot (0.8)^{11} \approx 0.0859

STEP 6

P(X=1)=(111)(0.2)1(0.8)100.2362P(X=1) = \binom{11}{1} \cdot (0.2)^1 \cdot (0.8)^{10} \approx 0.2362

STEP 7

P(X=2)=(112)(0.2)2(0.8)90.2953P(X=2) = \binom{11}{2} \cdot (0.2)^2 \cdot (0.8)^9 \approx 0.2953

STEP 8

P(X=3)=(113)(0.2)3(0.8)80.2215P(X=3) = \binom{11}{3} \cdot (0.2)^3 \cdot (0.8)^8 \approx 0.2215

STEP 9

Now, add these probabilities together to find P(X3)P(X \le 3): 0.0859+0.2362+0.2953+0.22150.83890.0859 + 0.2362 + 0.2953 + 0.2215 \approx 0.8389.

STEP 10

Since the total probability of all possible outcomes is 1, the probability of getting more than 3 right is 1P(X3)1 - P(X \le 3).
This is because either the student gets 3 or fewer questions right, or they get more than 3 right.
There are no other possibilities!

STEP 11

So, P(X>3)=10.83890.1611P(X > 3) = 1 - 0.8389 \approx 0.1611.

STEP 12

The probability of getting more than 3 questions right is approximately **0.1611**.

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