Where a = 30 and b = 55.
Probability (time = t) = 1/(b-a) = 1/ (55-30) = 1/25 = 0.04
b. P (t<45) = The area between t= 30 and t=45 minutes = (x-a) *1/(b-a) = (45-30) *0.04 = 0.6
c. P(t>48) = The area between t = 48 and t = 55 minutes = 1-((x-a) *1/(b-a)) = 1 – ((48 – 30) *0.04) = 1-(18*0.04) = 0.28
Question 2:
a. Mean = XXX of ages of all deer/ number XX XXXX = XXXX/XX = XX.XX
XXXXXXXX deviation of sample =
Therefore, = root of (((47-35.37) ^2+(XX-35.XX) ^X+&XXXXXX;. +(XX-35.37) ^2)/ (30-X))
= XX.XX
b. XXXXX XXXXXXX XXXXX XXXXXXX, XX can XXX XXXX XXX XXXX XX XXX population and standard XXXXXXXXX are:
XXXX = 35.37
XX = s/XXXX(n) = XX.53/ (root (XX)) = 1.922
XXXXXXXXXXX (age >=XX) = 1- probability (age &XX;= 30) = 1 – P (X&XX;= (x-mean)/SE)
= X – P (z&XX;= (30-35.XX)/1.XXX) = X – P (z<=-2.XX) = X – 0.002635 = 0.XXX
Question 3:
a. Since, XXX entire class XXXXX would follow a normal XXXXXXXXXXXX, therefore, there any XXXXXX of samples that are XXXXX XXXX XXX population will XXXXXX a normal distribution. XXXXXXXXX, the XXXXXXXX deviation XXX mean of XXX sample can XX used XX XXXXXXXXX XXX mean and XXXXXXXX deiviation of XXX XXXXXXXXXX scores of the population.
b. XXXX = 75.7
XXXXXXXX deviation = 24.X
XXXXXXXX = 24.X^X/n
XXXXXX variance = XX.255
24.3^2/n = 21.255 or n = XX.78 or 28
So, most likely XXXXX size would be 28
XXXXXXXX X:
Z score = (x-mean)/standard XXXXXXXXX
a. Z = (250-210)/ 50 = XX/50 = 0.8
X (XXXXX&XX;250) = 1- P(XXXXX&XX;250) = 1 – X(X<X.X) = 1- X.XXX = X.XXX
b. P(XXXXXX<235) = X (X < (XXX – 210)/50)) = X (z&XX; 0.X) = 0.XXX
XXXXXXXXXX of bulbs XXXX XXXXX to be XXXXXXXX within XXX XXXXX = XX.2%
c. P(XXX<XXXXXX<400) = P (lights&XX;XXX)-X(lights&XX;XXX) = X (z< (XXX-XXX)/50) – P(z < (XXX-210)/50)
= X(z&XX;X.X) – X (z< -0.X)
= X.999 – X.XXX = 0.578
Number of bulbs XXXX lifetime between XXX and XXX h = 57.8% XX XXXX = 1156
Question 5:
XXXX = X.X
Standard deviation = 0.8
N = 64
XXXXX N>30, we XXX apply XXXXXXX limit theorem,
Mean XX XX months XX find a job, XXXX = X.4
XX, Standard XXXXX = standard XXXXXXXXX/root(n) = 0.8/XXXX(XX) = 0.X/8 = X.1
95% confidence XXXXXXXX:
It XXX an XXXXXXXX XX XXXX- (z*XX) to XXXX + (X*XX)
XX% = X.XX XX the cumulative XXXXXXXXXXX XXXXX leaves X.05/X on XXXXXX side of the interval = X.XXX
X (b< z < a) = X.95
P (X&XX;a) – X(X&XX;b) = X.XX
Or, X(X&XX; b) = 0.025
XX, b = -X.XX
Therefore, XX XXXXXXXX a = X.96
Number of months within XXXXX a graduate will get a job XXXX XX% XXXXXXXXXX:
XXXXX limit = mean – (X*XX) = 5.X – (X.96*X.X) = X.X – X.XXX = 5.XXX
Upper limit = mean + (X*XX) = X.X + (1.96*0.1) = 5.4 + X.196 = 5.XXX
Question X:
a. Margin of XXXXX = X%
b. XXXXXXXXXX interval:
X(b&XX;readers&XX;a) = P(z&XX;a)-X(X<b) = 0.XX
X(z<b) = X.XXX
b = -1.96
XX symmetry, a = X.96
SE = X.XX XXXX = X.XX Therefore, XXX confidence interval is ( 0.63 (+/-) 1.96*0.05)
XX, X.XX-0.098 XX X.XX+X.XXX
XX, X.XXX to X.XXX
XXXXXXXXX, XX can say XX with 95% confidence XXXX XX XXXXX 53.X% XXX XX XXX 72.X% XX the XXXXXX regularly read XX XXXXX a XXXX XX XXX newspaper
c. XXXXXXXXXX XXXXX = XX%