Probability and Statistics Tutorial 3: week 4 Problem 6A () > - - PowerPoint PPT Presentation

Probability and Statistics

Tutorial 3: week 4

Problem 6A

• 𝑈

1 ∼ 𝐹𝑦𝑞 0.2

• 𝑈2~ 𝐹𝑦𝑞 0.25
• 𝑈

1 𝑏𝑜𝑒 𝑈2 𝑏𝑠𝑓 𝑗𝑜𝑒𝑓𝑞𝑓𝑜𝑒𝑓𝑜𝑢

(𝐛) 𝑸 𝑼𝟐 > 𝟒 𝒏𝒋𝒐𝒗𝒖𝒇𝒕 ANSWER:

• Use the cumulative distribution for 𝑈

1: 𝐺(𝒖)

• 𝑄 𝑈

1 > 3 = 1 − 𝑄(𝑼𝟐 ≤ 𝟒) = 1 – 𝐺(3)

• Analytically:

𝐺 𝑢 = 1 − 𝑓−𝜇𝑦 = 1 − 1 − 𝑓−0.2×3 = 0.54881 R code: 1 – pexp(t,rate) 1- pexp(3,0.2)

Problem 6B

• 𝑈

1 ∼ 𝐹𝑦𝑞 0.2

• 𝑈2~ 𝐹𝑦𝑞 0.25
• 𝑈

1 𝑏𝑜𝑒 𝑈 2 𝑏𝑠𝑓 𝑗𝑜𝑒𝑓𝑞𝑓𝑜𝑒𝑓𝑜𝑢

(𝐜) 𝑸 𝑼𝟑 < 𝟓 𝒏𝒋𝒐𝒗𝒖𝒇𝒕 ANSWER:

• Use the cumulative distribution for 𝑈

1: 𝐺(𝒖)

• 𝑄 𝑈

2 < 4 = 𝑄(𝑈 2 ≤ 𝟓) = 1 – 𝐺(4)

(WHY?)

• Analytically:

𝐺 𝑢 = 1 − 𝑓−𝜇2𝑦 = 1 − 𝑓−0.25×4 = 0.632 R code: pexp(t,rate) pexp(4,rate = 0.25)

Normal Distribution

• Let R be a random variable

measuring the reading speed of students

• Bell shaped;
• Symmetric about the mean 𝜈
• Shape (height and width) given by

the standard deviation: 𝜏

• Normally given by: 𝑄 𝑌 ≤ 𝑏 =

F 𝑏

• For calculations, we use the

standard Normal curve

• 𝑎 ∼ 𝑂(0, 1)

𝑼𝒊𝒇 𝑶𝒑𝒔𝒏𝒃𝒎 𝑬𝒋𝒕𝒖𝒔𝒋𝒄𝒗𝒖𝒋𝒑𝒐

Problem 7A

• Let R be a random variable measuring the

reading speed of students

• What is the distribution of R?

𝑆 ∼ 𝑂(125, 242) 𝑆 ∼ 𝑂 125, 242 → 𝑎 ∼ 𝑂(0, 1) 𝑎 = 𝑌 − 𝜈 𝜏 𝑎 = 𝑌 − 125 24 𝒃 𝑸 𝑺 < 𝟐𝟏𝟏 ? ANSWER:

• First, we standardize. What does this mean?

𝑆 ∼ 𝑂 125, 242 → 𝑎 ∼ 𝑂(0, 1)

(WHY?)

𝑄 𝑆 < 100 = 𝑄 𝑆 ≤ 100 = 𝑄 𝑎 < 100−125

24

= 𝐺

100−125 24

= Φ

100−125 24

Use R 𝑞𝑜𝑝𝑠𝑛(100−125

24

, 𝑛𝑓𝑏𝑜 = 0, 𝑡𝑒 = 1) = 0.1488 Or Skip standardization : 𝑞𝑜𝑝𝑠𝑛(100, 𝑛𝑓𝑏𝑜 = 125, 𝑡𝑒 = 24)

Problem 7b

• Let R be a random variable measuring the

reading speed of students

• What is the distribution of R?

𝑆 ∼ 𝑂(125, 242) 𝑆 ∼ 𝑂 125, 242 → 𝑎 ∼ 𝑂(0, 1) 𝑎 = 𝑌 − 𝜈 𝜏 𝑎 = 𝑌 − 125 24 𝒄 𝑸 𝑺 > 𝟐𝟓𝟏 ? ANSWER: 𝑄 𝑆 > 140 = 1 − 𝑄 𝑆 ≤ 140 Skipping standardization: 1 − 𝑞𝑜𝑝𝑠𝑛(140, 𝑛𝑓𝑏𝑜 = 125, 𝑡𝑒 = 24) = 0.2660

Problem 7c

• Let R be a random variable measuring

the reading speed of students

• What is the distribution of R?

𝑆 ∼ 𝑂(125, 242) 𝑆 ∼ 𝑂 125, 242 → 𝑎 ∼ 𝑂(0, 1) 𝑎 = 𝑌 − 𝜈 𝜏 𝑎 = 𝑌 − 125 24 𝒅 𝑸 𝟐𝟐𝟏 ≤ 𝑺 ≤ 𝟐𝟒𝟏 ? ANSWER: 𝑸 𝟐𝟐𝟏 ≤ 𝑺 ≤ 𝟐𝟒𝟏 = F 130 − F(110) Skipping standardization:

𝑞𝑜𝑝𝑠𝑛(130, 𝑛𝑓𝑏𝑜 = 125, 𝑡𝑒 = 24) - 𝑞𝑜𝑝𝑠𝑛 110, 𝑛𝑓𝑏𝑜 = 125, 𝑡𝑒 = 24 = 0.317

Problem 7d

• Let R be a random variable measuring

the reading speed of students

• What is the distribution of R?

𝑆 ∼ 𝑂(125, 242) 𝑆 ∼ 𝑂 125, 242 → 𝑎 ∼ 𝑂(0, 1) 𝑎 = 𝑌 − 𝜈 𝜏 𝑎 = 𝑌 − 125 24 𝒄 𝑸(𝑺 > 𝟑𝟏𝟏)? ANSWER: 𝑸 𝑺 > 𝟑𝟏𝟏 = 𝟐 − 𝑸 𝑺 ≤ 𝟑𝟏𝟏 = 1 − F(200) Skipping standardization:

1 - 𝑞𝑜𝑝𝑠𝑛 200, 𝑛𝑓𝑏𝑜 = 125, 𝑡𝑒 = 24 = 0.00089

Problem 8A and B

• 𝑌1 ∼ 𝐹𝑦𝑞 𝜇1
• 𝑌2~ 𝐹𝑦𝑞 𝜇2
• 𝑌1 𝑏𝑜𝑒 𝑌2 𝑏𝑠𝑓 𝑗𝑜𝑒𝑓𝑞𝑓𝑜𝑒𝑓𝑜𝑢

𝑸(𝒀𝟐 > 𝒃) Hint: use the analytical form of the cdf: 𝐺 𝑏 = 1 − 𝑓𝜇1𝑏 ANSWER:

• Analytically:

𝑄 𝑌1 > 𝑏 = 1 − 𝑄 𝑌 ≤ 𝑏 = 1 − 𝐺(𝑏) = 1 − 1 − 𝑓−𝜇1𝑏 = 𝑓−𝜇1𝑏 (same for B but with 𝜇2

Problem 8c

• 𝑌1 ∼ 𝐹𝑦𝑞 𝜇1
• 𝑌2~ 𝐹𝑦𝑞 𝜇2
• 𝑌1 𝑏𝑜𝑒 𝑌2 𝑏𝑠𝑓 𝑗𝑜𝑒𝑓𝑞𝑓𝑜𝑒𝑓𝑜𝑢

𝒂 = 𝒏𝒋𝒐 𝒀𝟐, 𝒀𝟑 𝑸 𝒂 > 𝒃 ? Hint: if Z is minimum of 𝒀𝟐𝒃𝒐𝒆 𝒀𝟑, then Z can only be larger than some ‘a’ if……. (THINK): ANSWER:

• …if both 𝑌1𝑏𝑜𝑒 𝑌2 are larger than a.
• 𝑄 𝑎 > 𝑏 = 𝑄(𝑌1 > 𝑏 ∩ 𝑌2 > 𝑏)

= P X1 > a × 𝑄 𝑌2 > 𝑏 (ind.) = 𝑓−𝜇1𝑏𝑓−𝜇2𝑏 = 𝑓−𝑏(𝜇1+𝜇2)

Problem 8d

• 𝑌1 ∼ 𝐹𝑦𝑞 𝜇1
• 𝑌2~ 𝐹𝑦𝑞 𝜇2
• 𝑌1 𝑏𝑜𝑒 𝑌2 𝑏𝑠𝑓 𝑗𝑜𝑒𝑓𝑞𝑓𝑜𝑒𝑓𝑜𝑢

𝑫𝑬𝑮 𝒑𝒈 𝒂 𝑮 𝒃 = 𝑄 𝑎 ≤ 𝑏 = 1 − 𝑄 𝑎 > 𝑏 = 1 − 𝑓−𝑏(𝜇1+𝜇2) PDF of Z How do we move from cdf to pdf? ANSWER: differentiation!!!!! 𝑔 𝑏 = 𝑒𝐺 𝑏 𝑒𝑏

𝑔 𝑏 = 𝜇1 + 𝜇2 𝑓−𝑏(𝜇1+𝜇2) 𝒂 ∼ 𝐟𝐲𝐪 𝝁𝟐 + 𝝁𝟑

Problem 6C (l (later)

• 𝑈

1 ∼ 𝐹𝑦𝑞 0.2

• 𝑈2~ 𝐹𝑦𝑞 0.25
• 𝑈

1 𝑏𝑜𝑒 𝑈2 𝑏𝑠𝑓 𝑗𝑜𝑒𝑓𝑞𝑓𝑜𝑒𝑓𝑜𝑢

𝐜 𝑼𝒏: 𝒙𝒃𝒋𝒖𝒋𝒐𝒉 𝒖𝒋𝒏𝒇 𝒗𝒐𝒖𝒋𝒎 𝒇𝒋𝒖𝒊𝒇𝒔 𝒄𝒗𝒕 𝟐 𝒑𝒔 𝒄𝒗𝒕 𝟑: ANSWER: 𝑼𝒏 = 𝒏𝒋𝒐 𝑼𝟐, 𝑼𝟑 What is is th the dis istribution of

• f Tm?

𝑼𝒏 ∼ 𝒇𝒚𝒒(𝝁𝟐 + 𝝁𝟑)

• 𝑄 𝑈𝑛 < 2 = 1 − e2 0.45 = 0.593

(WHY?)

THE END

• ANY QUESTIONS??