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2.4.13 Model with probability distributions

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Question 25

A botanist observes that the number of seeds, $S$ , that germinate in a large nursery plot follows a binomial distribution $\sim \text{B}(n, p)$ . The expected number of germinated seeds is 200.

State the variance of $S$ in terms of $p$ .

[1]

A normal distribution is used as an approximation for $S$ . Using this approximation and a continuity correction, the probability that 180 or fewer seeds germinate is approximately 0.0384.

Show that $200−200p=11.02\sqrt{200 - 200p} = 11.02$ to 4 significant figures.

[5]

Hence find the value of $p$ , giving your answer to 2 significant figures.

[3]

Markscheme

2.4.13 Model with probability distributions Questions

A Level
Maths
2.4.13 Model with probability distributions

67 exam-style questions on OCR (MEI) A Level Maths 2.4.13 Model with probability distributions. Each one has a worked solution and a mark scheme showing where the marks go.

2.4.13 Model with probability distributions

Question bank

2.4.13 Model with probability distributions

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Question 25

A botanist observes that the number of seeds, $S$ , that germinate in a large nursery plot follows a binomial distribution $\sim \text{B}(n, p)$ . The expected number of germinated seeds is 200.

State the variance of $S$ in terms of $p$ .

[1]

A normal distribution is used as an approximation for $S$ . Using this approximation and a continuity correction, the probability that 180 or fewer seeds germinate is approximately 0.0384.

Show that $200−200p=11.02\sqrt{200 - 200p} = 11.02$ to 4 significant figures.

[5]

Hence find the value of $p$ , giving your answer to 2 significant figures.

[3]

Markscheme

2.4.13 Model with probability distributions Questions

A Level
Maths
2.4.13 Model with probability distributions

67 exam-style questions on OCR (MEI) A Level Maths 2.4.13 Model with probability distributions. Each one has a worked solution and a mark scheme showing where the marks go.