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 XX XXX XXXX/ XXXXXX of deer = 1061/30 = XX.XX
XXXXXXXX deviation XX sample =
Therefore, = root of (((XX-XX.XX) ^2+(25-35.37) ^2+&XXXXXX;. +(64-35.37) ^2)/ (30-1))
= 10.XX
b. XXXXX Central XXXXX theorem, XX can XXX that XXX XXXX XX the population XXX standard deviation XXX:
Mean = XX.XX
XX = s/root(n) = 10.53/ (XXXX (XX)) = X.922
XXXXXXXXXXX (age &XX;=30) = 1- probability (age <= XX) = X – X (z<= (x-mean)/SE)
= X – P (z&XX;= (30-35.XX)/X.XXX) = 1 – P (z&XX;=-2.79) = X – 0.002635 = X.997
XXXXXXXX X:
a. XXXXX, XXX entire XXXXX marks would XXXXXX a XXXXXX distribution, XXXXXXXXX, XXXXX XXX number XX samples that are drawn from the population will follow a XXXXXX distribution. Therefore, XXX XXXXXXXX deviation XXX mean of the XXXXXX can be used XX XXXXXXXXX XXX mean XXX XXXXXXXX deiviation XX the XXXXXXXXXX XXXXXX of XXX XXXXXXXXXX.
b. Mean = XX.7
Standard deviation = XX.3
XXXXXXXX = 24.3^X/n
Sample variance = XX.XXX
XX.X^2/n = 21.XXX or n = XX.XX or 28
XX, XXXX likely class size would XX XX
Question 4:
Z XXXXX = (x-XXXX)/standard XXXXXXXXX
a. X = (250-XXX)/ 50 = XX/50 = X.8
P (light>XXX) = X- P(light<250) = X – X(X&XX;X.X) = X- 0.788 = 0.212
b. X(XXXXXX<235) = P (z < (XXX – XXX)/XX)) = P (X&XX; 0.5) = X.XXX
XXXXXXXXXX XX bulbs that XXXXX to be XXXXXXXX XXXXXX XXX XXXXX = XX.2%
c. X(XXX&XX;lights<XXX) = P (XXXXXX&XX;XXX)-P(XXXXXX&XX;200) = P (X&XX; (400-XXX)/XX) – P(X < (XXX-210)/XX)
= P(X&XX;X.8) – X (z&XX; -X.X)
= 0.XXX – 0.XXX = 0.XXX
XXXXXX XX bulbs XXXX lifetime XXXXXXX 200 and 400 h = XX.8% of 2000 = XXXX
Mean = X.X
Standard deviation = X.8
N = 64
Since N>30, we XXX XXXXX XXXXXXX XXXXX XXXXXXX,
XXXX no of XXXXXX to XXXX a XXX, XXXX = 5.X
XX, XXXXXXXX error = standard XXXXXXXXX/XXXX(n) = 0.8/XXXX(XX) = 0.8/8 = X.X
XX% confidence interval:
XX XXX an XXXXXXXX of XXXX- (X*SE) to mean + (z*SE)
XX% = X.XX is XXX cumulative probability XXXXX XXXXXX X.05/2 on either side of the XXXXXXXX = 0.XXX
P (b&XX; X &XX; a) = 0.95
P (X&XX;a) – P(z<b) = 0.95
XX, P(X< b) = X.XXX
Or, b = -X.96
Therefore, XX symmetry a = X.XX
XXXXXX XX months XXXXXX which a XXXXXXXX XXXX XXX a XXX XXXX 95% confidence:
XXXXX XXXXX = XXXX – (z*SE) = 5.X – (1.XX*X.X) = 5.4 – 0.XXX = X.204
XXXXX XXXXX = XXXX + (X*SE) = 5.4 + (X.XX*X.1) = 5.X + 0.196 = X.596
a. Margin of XXXXX = 5%
b. Confidence interval:
P(b&XX;XXXXXXX&XX;a) = X(z&XX;a)-X(X&XX;b) = X.95
X(X&XX;b) = X.025
b = -X.XX
By XXXXXXXX, a = 1.XX
XX = X.XX Mean = X.XX XXXXXXXXX, the confidence interval is ( 0.XX (+/-) 1.XX*X.XX)
XX, 0.XX-0.XXX XX 0.63+0.098
XX, 0.XXX to 0.728
Therefore, we XXX XXX XX XXXX XX% XXXXXXXXXX XXXX at least 53.2% XXX at max XX.X% XX the XXXXXX regularly read at least a XXXX XX the XXXXXXXXX
c. XXXXXXXXXX XXXXX = XX%