SPM.2.SL.TZ0.8

pestleMathematicsAASLPaper 2SPM· sl-4-6-combined-mutually-exclusive-conditional-independence-prob-diagramssource ↗

The length, X mm, of a certain species of seashell is normally distributed with mean 25 and variance, ${\sigma ^2}$ .

The probability that X is less than 24.15 is 0.1446.

A random sample of 10 seashells is collected on a beach. Let Y represent the number of seashells with lengths greater than 26 mm.

Find P(24.15 < X < 25).

[2]

Find $\sigma$ , the standard deviation of X.

[3]

b.i.

Hence, find the probability that a seashell selected at random has a length greater than 26 mm.

[2]

b.ii.

Find E(Y).

[3]

Find the probability that exactly three of these seashells have a length greater than 26 mm.

[2]

A seashell selected at random has a length less than 26 mm.

Find the probability that its length is between 24.15 mm and 25 mm.

[3]

Markscheme / solution

attempt to use the symmetry of the normal curve (M1)

eg diagram, 0.5 − 0.1446

P(24.15 < X < 25) = 0.3554 A1

[2 marks]

use of inverse normal to find z score (M1)

z = −1.0598

correct substitution $\frac{{24.15 - 25}}{\sigma } = - 1.0598$ (A1)

$\sigma$ = 0.802 A1

[3 marks]

b.i.

P(X > 26) = 0.106 (M1)A1

[2 marks]

b.ii.

recognizing binomial probability (M1)

E(Y) = 10 × 0.10621 (A1)

= 1.06 A1

[3 marks]

P(Y = 3) (M1)

= 0.0655 A1

[2 marks]

recognizing conditional probability (M1)

correct substitution A1

$\frac{{0.3554}}{{1 - 0.10621}}$

= 0.398 A1

[3 marks]

Examiners’ report

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b.i.

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b.ii.

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