Other
Statistics_Homework
- Course
- Institution
[Show more]
Seller
FollowDuryoDana
Content preview
NOTE : SUBMITTED ANSWERS ON TABLES 4,6,7 FOR 1-3 QUESTIONS & TABLE 5 FOR 4-6 QueQ 1)
If one Pearson is selected at random from the population in Table 4, calculate the probability of a bl
Would you consider this result unusual – why or why not?
Thumb rule to decide is looking at a data histogram a good initial way to do. If the points collected in
normal distribution. It hardly shows as below
TABLE – 4
P(X > 136) = (136-120)/9.75
= 16/9.75 = 1.641
= P ( Z >1.641) From Standard Normal Table
= 0.0504
Result is usual as it lies in [0,1]
TABLE – 6
Mean of Sample 24 observations ( u ) = 113.85
Standard Deviation ( sd )=12.76
Normal Distribution = Z= X- u / sd ~ N(0,1)
P(X > 136) = (136-113.85)/12.76
= 22.15/12.76 = 1.7359
= P ( Z >1.736) From Standard Normal Table
= 0.0413
Result is usual as it lies in [0,1]
TABLE – 7
,If five people are selected at random, calculate the probability that their mean blood pressure is gr
this result unusual – why or why not?
TABLE – 4
P(X > 136) = (136-120)/9.75/ Sqrt ( 5 )
= 16/4.36= 3.6694
= P ( Z >3.6694) From Standard Normal Table
= 0.0001
Result is unusual as it lies almost in extremes[0,1]
TABLE – 6
P(X > 136) = (136-113.85)/12.76/ Sqrt ( 5 )
= 22.15/5.706= 3.8816
= P ( Z >3.8816) From Standard Normal Table
= 0.0001
Result is unusual as it lies almost in extremes[0,1]
TABLE – 7
P(X > 136) = (136-113.5)/11.12/ Sqrt ( 5 )
= 22.5/4.973= 4.5244
= P ( Z >4.5244) From Standard Normal Table
=0
Result is unusual as it lies at extremes [0,1]
Q 3) If five people are selected at random, calculate the probability that their mean blood pressure
TABLE – 4
To find P(a <= Z <=b) = F(b) - F(a)
, TABLE – 6
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 130) = (130-113.85)/12.76/ Sqrt ( 5 )
= 16.15/5.7064
= 2.8301
= P ( Z <2.8301) From Standard Normal Table
= 0.99767
P(X < 136) = (136-113.85)/12.76/ Sqrt ( 5 )
= 22.15/5.7064 = 3.8816
= P ( Z <3.8816) From Standard Normal Table
= 0.99995
P(130 < X < 136) = 0.99995-0.99767
= 0.0023
TABLE – 7
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 130) = (130-113.5)/11.12/ Sqrt ( 5 )
= 16.5/4.973
= 3.3179
= P ( Z <3.3179) From Standard Normal Table
= 0.99955
P(X < 136) = (136-113.5)/11.12/ Sqrt ( 5 )
= 22.5/4.973 = 4.5244
= P ( Z <4.5244) From Standard Normal Table
=1
P(130 < X < 136) = 1-0.99955
= 0.0005
The benefits of buying summaries with Stuvia:
Guaranteed quality through customer reviews
Stuvia customers have reviewed more than 700,000 summaries. This how you know that you are buying the best documents.
Quick and easy check-out
You can quickly pay through credit card or Stuvia-credit for the summaries. There is no membership needed.
Focus on what matters
Your fellow students write the study notes themselves, which is why the documents are always reliable and up-to-date. This ensures you quickly get to the core!
Frequently asked questions
What do I get when I buy this document?
You get a PDF, available immediately after your purchase. The purchased document is accessible anytime, anywhere and indefinitely through your profile.
Satisfaction guarantee: how does it work?
Our satisfaction guarantee ensures that you always find a study document that suits you well. You fill out a form, and our customer service team takes care of the rest.
Who am I buying these notes from?
Stuvia is a marketplace, so you are not buying this document from us, but from seller DuryoDana. Stuvia facilitates payment to the seller.
Will I be stuck with a subscription?
No, you only buy these notes for $3.93. You're not tied to anything after your purchase.
Can Stuvia be trusted?
4.6 stars on Google & Trustpilot (+1000 reviews)
52355 documents were sold in the last 30 days
Founded in 2010, the go-to place to buy study notes for 14 years now