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notes 4.11 regarding the, Quizzes of Mathematical Analysis

math notes about hte staticitcs chapter

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2024/2025

Uploaded on 04/27/2025

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Math 2810 Course Notes 2/28
4.11. The Central Limit Theorem
Theorem 1. (Central Limit Theorem) If X1, X2, . . . , Xnis a large (n > 30) sample from a
population with mean µand variance σ2, then
1. (Sum) SnN(nµ, 2), where Sn=X1+X2+· ·· +Xn.
2. (Average) X=Sn
nNµ, σ2
n.
Example 2. Suppose ages of students in a university have population mean 22.3 years and
standard deviation 4 years. Estimate the probability that the average age of 64 randomly
selected students is greater than 23 years old.
Let Xbe the average age of the 64 students. According to the Central Limit Theorem,
XN22.3,42
64=N(22.3,(0.5)2).
So
P(X > 23) = PZ > 23 22.3
0.5=P(Z > 1.4) = 1 Φ(1.4).
Example 3. Estimate the probability that the sum of 100 fair die rolls is greater than 400.
Recall one dice roll has population mean µ=7
2= 3.5 and variance σ2=35
12 2.9167.
Let S100 be the sum of 100 dice rolls. According to the Central Limit Theorem,
S100 N(100 ·3.5,100 ·2.9167) = N(350,291.67).
Therefore
P(S100 >400) = PZ > 400 350
291.67 P(Z > 2.93) = 1 Φ(2.93).
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Math 2810 Course Notes 2/

4.11. The Central Limit Theorem

Theorem 1. (Central Limit Theorem) If X 1 , X 2 ,... , Xn is a large (n > 30 ) sample from a population with mean μ and variance σ^2 , then

  1. (Sum) Sn ≈ N (nμ, nσ^2 ), where Sn = X 1 + X 2 + · · · + Xn.
  2. (Average) X = Sn n

≈ N

μ, σ^2 n

Example 2. Suppose ages of students in a university have population mean 22.3 years and standard deviation 4 years. Estimate the probability that the average age of 64 randomly selected students is greater than 23 years old.

Let X be the average age of the 64 students. According to the Central Limit Theorem,

X ≈ N

= N (22. 3 , (0.5)^2 ).

So

P (X > 23) = P

Z >

= P (Z > 1 .4) = 1 − Φ(1.4).

Example 3. Estimate the probability that the sum of 100 fair die rolls is greater than 400.

Recall one dice roll has population mean μ =

= 3.5 and variance σ^2 =

Let S 100 be the sum of 100 dice rolls. According to the Central Limit Theorem,

S 100 ≈ N (100 · 3. 5 , 100 · 2 .9167) = N (350, 291 .67).

Therefore

P (S 100 > 400) = P

Z >

≈ P (Z > 2 .93) = 1 − Φ(2.93).

Example 4. Suppose the waiting time (in min) for a certain bus has distribution U (0, 20). Carol takes this bus 40 times every month.

  1. Estimate the probability that her monthly average waiting time per ride is shorter than 9 minutes.

Recall that U (0, 20) has mean μ =

= 10 and σ^2 =

(20 − 0)^2

. Let X be her monthly average waiting time per ride. According to the Central Limit Theorem,

X ≈ N

= N

Therefore

P (X < 9) = P

Z < 9 q−^10 5 6

 (^) q−^1 5 6

  1. Estimate the probability that her monthly total waiting time is shorter than 6 hours.

Notice that 6 hours total in 40 rides is equivalent to 9 minutes per ride on average. So the answer is the same as part 1.

Normal Approximation to Binomial

If X ∼ Bin(n, p), and np > 10 , n(1 − p) > 10, then

  1. (Number of Successes) X ≈ N (np, np(1 − p))
  2. (Estimator for p, won’t use until 5.3)

pˆ =

X

n

≈ N

p, p(1 − p) n

At-Home Lecture Review

  1. A Bernoulli trial with success rate 80% was conducted 100 times.

(a) Find the exact value of the probability of at least 90 successes. 0. 0059696 (b) Estimate the probability of at least 90 successes. 0. 0087

  1. Abby rolls a fair six-sided die 36 times. Mark rolls a fair six-sided die 35 times. Do not use continuity correction in this problem. (a) Estimate the probability that the sum of Alice’s rolls is less than 120. 0. 2776 (b) Estimate the probability that the sum of Alice’s rolls and the sum of Mark’s rolls are both less than 120. 0. 1114 (c) Estimate the 60th percentile of the sum of Mark’s rolls. 125. 026 (d) Estimate the probability that the sum of Alice’s rolls is less than the sum of Mark’s rolls. 0. 4052 (e) Estimate the probability that Mark rolls more than five 1s. 0. 6580 (f) Estimate the probability that Alice rolls more 1s than Mark. 0. 5199 (g) Estimate the probability that Alice and Mark together roll more than ten 1s in total. 0. 7190