HW11
Name: Riya Rathod
Subject: Mathematical Foundation Of Machine Learning
Question 1
Answer:
X ∼ N (µ = 46.8, σ = 1.75).
x−µ
z=
σ
(a) Probability that the speed is at most 50 km/h
50 − 46.8
= 1.8286.
1.75
Using a standard normal table:
z=
P (Z ≤ 1.83) ≈ 0.9664.
So the probability is about 96.64%.
(b) Probability that the speed is at least 48 km/h
z=
48 − 46.8
= 0.6857.
1.75
P (Z ≤ 0.686) ≈ 0.754.
So:
P (X ≥ 48) = 1 − 0.754 = 0.246.
This means about 24.6% of mopeds go at least 48 km/h.
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(c) Probability that speed is within 1.5 standard deviations of the mean
P (|X − µ| ≤ 1.5σ) = P (−1.5 ≤ Z ≤ 1.5).
Φ(1.5) = 0.9332,
Φ(−1.5) = 0.0668.
P = 0.9332 − 0.0668
P = 0.8664
So the probability is about 86.64%.
Question 2
Tree diameters follow:
X ∼ N (8.8, 2.8).
(a) Probability that diameter is at least (or exceeds) 10
inches
z=
10 − 8.8
= 0.4286.
2.8
P (Z ≤ 0.429) = 0.666.
P (X ≥ 10) = 1 − 0.666
P (X ≥ 10) = 0.334
So the probability is 33.4%.
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(b) Probability that diameter exceeds 20 inches
z=
20 − 8.8
= 4.
2.8
P (Z > 4) ≈ 0.00003
So almost 0% of trees exceed 20 inches.
(c) Probability that diameter is between 5 and 10 inches
For 5 inch:
z=
5 − 8.8
= −1.357,
2.8
Φ(−1.357) = 0.087.
For 10 inch:
Φ(0.4286) = 0.666.
P = 0.666 − 0.087
P = 0.579
So about 57.9% of trees are between 5–10 inches wide.
(d) Find c such that µ ± c contains 98% of values
For 98%, the z value is
z = 2.326.
c = zσ = 2.326 × 2.8
c = 6.513
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(e) Probability that at least one of four trees exceeds
10 in
From earlier:
p = 0.334.
Probability that a tree does not exceed 10:
1 − p = 0.666.
All 4 fail to exceed 10:
(0.666)4 = 0.196.
At least one exceeds:
1 − 0.196 = 0.804.
So the probability is 80.4%.
Question 3
Birth weight distribution:
X ∼ N (3432, 482).
(a) Probability that weight exceeds 4000 g; and between
3000 and 4000 g
For 4000 g:
z = 1.178,
Φ(1.178) = 0.880.
P (X > 4000) = 1 − 0.880 = 0.120.
For 3000 g:
z = −0.897,
Φ(−0.897) = 0.185.
Between 3000–4000:
0.880 − 0.185 = 0.695.
So that probability is 69.5%.
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(b) Probability that weight is < 2000 g or > 5000 g
2000 g:
P = 0.0015.
5000 g:
P = 0.0006.
Total:
0.0015 + 0.0006 = 0.0021.
So the probability is 0.21%.
(c) Probability that weight exceeds 7 lb
Converting:
7 lb = 3175 g.
z = −0.533.
P (X > 3175) = 1 − Φ(−0.533) = 1 − 0.297
P (X > 3175) = 0.703
So about 70.3% of babies exceed 7 lb.
(d) Most extreme 0.1% of birth weights
z = ±3.09.
Low end : 3432 − 3.09(482) = 1942 g.
High end : 3432 + 3.09(482) = 4922 g.
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(e) Convert distribution to pounds and recompute
Y = 0.00220462X.
Mean:
µY = 7.56 lb.
Std dev:
σY = 1.06 lb.
z=
7 − 7.56
= −0.528.
1.06
P (Y > 7) = 1 − Φ(−0.528)
P (Y > 7) = 0.701
This is almost the same as before (0.703).
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