d 20 points Suppose that X n the wrights of sqsuh ina fars a

d. (20 points) Suppose that X n the wrights of sqsuh ina fars, and it is somnally farus, and it is sonnally distributed with mean -6 ll), niul variance r? \" 4 lh. (a) (G points) What s the probability that the seussh weigh at least 10 pound h at leasa 10 pounde? (0) (6 polats) What is the probobility that the sqjuash weigh between 4 and s pouds? (e) (S points) Fiad the value of c such that P(X S e-067.

Solution

t Test For Significance of Different Means
Set Up Hypothesis
Null Hypothesis, There Is NoSignificance between them Ho: u1 = u2
Alternative Hypothesis, There Is Significance between themH1: u1 != u2
Test Statistic
X (Mean)=10; Standard Deviation (s.d1)=4
Number(n1)=15
Y(Mean)=12; Standard Deviation(s.d2)=2
Number(n2)=15

we use Test Statistic (t) = (X-Y)/Sqrt(S^2(1/n1+1/n2))
to=10-12/Sqrt((10( 1 /15+ 1/15 ))
to=-2/1.1547
to=-1.7321
| to | =1.7321
Critical Value
The Value of |t a| with (n1+n2-2) i.e 28 d.f is 2.048
We got |to| = 1.7321 & | t a | = 2.048
Make Decision
Hence Value of |to | < | t a | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) - Ha : ( P != -1.7321 ) = 0.0939
Hence Value of P0.05 < 0.0939,Here We Do not Reject Ho

Normal Distribution
Mean ( u ) =6
Standard Deviation ( sd )=2
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
a)
P(X >= 10) = (10-6)/2
= 4/2 = 2
= P ( Z >2) From Standard Normal Table
= 0.0228                  

b)
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 8) = (8-6)/2
= 2/2 = 1
= P ( Z <1) From Standard Normal Table
= 0.84134
P(X < 2.15) = (2.15-6)/2
= -3.85/2 = -1.925
= P ( Z <-1.925) From Standard Normal Table
= 0.02711
P(8 < X < 2.15) = 0.02711-0.84134 = -0.8142                  

c)
P ( Z < x ) = 0.67
Value of z to the cumulative probability of 0.67 from normal table is 0.44
P( x-u/s.d < x - 6/2 ) = 0.67
That is, ( x - 6/2 ) = 0.44
--> x = 0.44 * 2 + 6 = 6.88