Statistics_Homework

NOTE : SUBMITTED ANSWERS ON TABLES 4,6,7 FOR 1-3 QUESTIONS & TABLE 5 FOR 4-6 Que

Q 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