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로그인 회원가입 You are solving a Korean CSAT mathematics problem. Reason carefully, but the final response must contain exactly one final answer inside \boxed{}.

For multiple-choice problems, put only one of ①, ②, ③, ④, ⑤ inside \boxed{}. For short-answer problems, put only the final integer inside \boxed{}. Do not use external tools.

Problem 29 Letabe a natural number witha≤6. Using one die and one coin, perform the following experiment. Roll the die once. – If the outcome is less than or equal toa, toss the coin 5 times and record the number of heads. – If the outcome is greater thana, toss the coin 3 times and record the number of heads. Repeat this experiment 19200 times, and letXbe the number of times the recorded value is 3. Suppose that E(X) = 4800. Using the standard normal table on the right, compute P(X≤4800 + 30a), and let this value bek. Compute 1000×k. [4 points] z P(0≤Z≤z) 0.5 0.191 1.0 0.341 1.5 0.433 2.0 0.477 2.5 0.494 3.0 0.499 12 더보기

Let me solve this step by step.

Step 1: Find the probability that the recorded value is 3 in one trial.

Let 𝑝 p be the probability that the recorded value equals 3 in a single experiment.

𝑝

𝑃 ( die ≤ 𝑎 ) ⋅ 𝑃 ( 3  heads in  5  tosses ) + 𝑃 ( die

𝑎 ) ⋅ 𝑃 ( 3  heads in  3  tosses ) p=P(die≤a)⋅P(3 heads in 5 tosses)+P(die>a)⋅P(3 heads in 3 tosses)

𝑝

𝑎 6 ⋅ ( 5 3 ) ( 1 2 ) 5 + 6 − 𝑎 6 ⋅ ( 3 3 ) ( 1 2 ) 3 p= 6 a ​

⋅( 3 5 ​

)( 2 1 ​

) 5 + 6 6−a ​

⋅( 3 3 ​

)( 2 1 ​

) 3

𝑝

𝑎 6 ⋅ 10 32 + 6 − 𝑎 6 ⋅ 1 8

5 𝑎 96 + 6 − 𝑎 48

5 𝑎 + 2 ( 6 − 𝑎 ) 96

3 𝑎 + 12 96

𝑎 + 4 32 p= 6 a ​

⋅ 32 10 ​

6 6−a ​

⋅ 8 1 ​

= 96 5a ​

48 6−a ​

= 96 5a+2(6−a) ​

= 96 3a+12 ​

= 32 a+4 ​

Step 2: Use 𝐸 ( 𝑋 )

4800 E(X)=4800 to find 𝑎 a.

𝐸 ( 𝑋 )

19200 ⋅ 𝑝

19200 ⋅ 𝑎 + 4 32

600 ( 𝑎 + 4 )

4800 E(X)=19200⋅p=19200⋅ 32 a+4 ​

=600(a+4)=4800

𝑎 + 4

8    ⟹    𝑎

4 a+4=8⟹a=4

Step 3: Approximate 𝑃 ( 𝑋 ≤ 4800 + 30 𝑎 ) P(X≤4800+30a).

With 𝑎

4 a=4: 𝑝

4 + 4 32

1 4 p= 32 4+4 ​

= 4 1 ​

.

𝑋 ∼ Binomial ( 19200 , 1 4 ) X∼Binomial(19200, 4 1 ​

).

𝐸 ( 𝑋 )

19200 ⋅ 1 4

4800 E(X)=19200⋅ 4 1 ​

=4800

Var ( 𝑋 )

19200 ⋅ 1 4 ⋅ 3 4

3600 , 𝜎

60 Var(X)=19200⋅ 4 1 ​

⋅ 4 3 ​

=3600,σ=60

𝑃 ( 𝑋 ≤ 4800 + 30 ⋅ 4 )

𝑃 ( 𝑋 ≤ 4920 )

𝑃 ( 𝑋 − 4800 60 ≤ 120 60 )

𝑃 ( 𝑍 ≤ 2 ) P(X≤4800+30⋅4)=P(X≤4920)=P( 60 X−4800 ​

≤ 60 120 ​

)=P(Z≤2)

Using the table: 𝑃 ( 𝑍 ≤ 2 )

0.5 + 𝑃 ( 0 ≤ 𝑍 ≤ 2 )

0.5 + 0.477

0.977 P(Z≤2)=0.5+P(0≤Z≤2)=0.5+0.477=0.977.

So 𝑘

0.977 k=0.977, and 1000 𝑘

977 1000k=977.

977 977 ​

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