# Program Latih Tubi SPM Matematik Tambahan Kertas 2

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```					         TOPIC:

PROBABILITY DISTRIBUTIONS
1.

(a)      n = 8, p = 0.4, q = 0.6
(i) P(X = 2) = 8C2 (0.4)2 (0.6)6   √ A1

= 0.2090 √ A1
(ii) P(X > 2) = 1 – P(X ≤ 2)
√ M1    = 1 – [ P(X = 0) + P( x = 1) + P(X = 2) ]
= 1 – [ 8C0 (0.4)0 (0.6)8 + 8C1 (0.4)1 (0.6)7 +
8C (0.4)2 (0.6)6 ]
2                  √ M1
= 0.6846 √ A1
(b)       µ = 130, σ = 16                                 0.7357
√ M1
(i) P(114 < X < 150)                       0.1587
0.1056
= P (114 130  z  150  130)
16             16
= P(−1 < z < 1.25)
z
= 1 – P(z > 1) – P(z > 1.25)             −1    0   1.25

= 1 – 0.1587 – 0.1056 √ M1
= 0.7357     √ A1

(ii) P(X > 150) = P(z > 1.25) √ M1
= 0.1056
132
Total number of workers =               = 1250 √ A1
0.1056
2.
(a) Binomial Distribution
(i) p = 0.2, q = 0.8, n = 8
8        2     6   √ M1
P(x = 2) = C2 (0.2) (0.8)
= 0.294 / 0.2936     √ A1

√ M1
(ii) P(x < 3) = P(x= 0, 1, 3)
or = P(x=0) + P(x=1) + P(x=2)
= 0.797 / 0.7969      √ A1
(b) Normal Distributions
4.0  3.2                  0.9452                    0.0548
(i)   z               = 1.6   √ M1
0.5

z
P(X < 4) = P(Z < 1.6)                           0   1.6

√ M1
= 1 – P(Z > 1.6)

= 1 – 0.0548

= 0.945 / 0.9452        √ A1
(b) (ii) P(X > m) = 0.6                                         0.6
0.4
√ M1
m  3.2
P(Z >       0.5
) = 0.6                                  z
z   0

m  3.2
1 – P(Z > −       0.5
) = 0.4       OR        P(Z > z) = 0.6
1 – P(Z > −z) = 0.4
m  3.2
P(Z < −z) = 0.4
P(Z < −      0.5
) = 0.4                        z = −0.253

m  3.2
= − 0.253     √ M1
0.5

m = 3.07 // 3.073             √ A1
3.074 // 3.075
THE END
THANK YOU

```
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