A sample of size 20 from a normal distribution has sample mean 3.5 and sample variance 14.75. Find 90% confidence intervals for
1 answer:
Answer:
Variance =9.2983, 27.6927
Standard deviation = 3.0493, 5.2624
Step-by-step explanation:
n = 20
S = 14.75
X² = chisquare
N-1 = degree of freedom
Alpha = 0.1
Alpha/2 = 0.1/2 = 0.05
1-0.1/2 = 0.95
Using statistics table
X² 19,0.05 = 30.14
X² 19, 0.95 = 10.12
A. 19(14.75)/30.14 , 19(14 75)/10.12
= 9.2983, 27.6927 for the variance of the population
B. Standard deviation = square root of variance
= √9.2983, √27.6927
= 3.0493, 5.2624
Please use attachment for better understanding
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Answer:
y = x+6
Step-by-step explanation:
You can find an equation by using the 2-point form of the equation for a line:
y = (y2 -y1)/(x2 -x1)·(x -x1) +y1
Filling in the given values, we have ...
y = (9 -5)/(3 -(-1))·(x -(-1)) +5
y = (4/4)(x +1) +5 . . . . simplifying a bit
y = x +6
See the attached figure to better understand the problem
in the triangle ABD
tan 29=BD/(525+x)------> BD=(525+x)*tan 29------> equation 1
in the triangle CBD
tan 38=BD/x-------> BD=x*tan 38-----------> equation 2
equals 1 and 2
(525+x)*tan 29=x*tan 38----> 525*tan29+x*tan29=x*tan38
x*[tan38-tan29]=525*tan29---> x=525*tan29/[tan38-tan29]---> x=1282.12 ft
BD=x*tan 38-----> 1282.12*tan38-----> BD=1001.71 ft-----> 1002 ft
the answer is
1002 ft
in the triangle ABD
tan 29=BD/(525+x)------> BD=(525+x)*tan 29------> equation 1
in the triangle CBD
tan 38=BD/x-------> BD=x*tan 38-----------> equation 2
equals 1 and 2
(525+x)*tan 29=x*tan 38----> 525*tan29+x*tan29=x*tan38
x*[tan38-tan29]=525*tan29---> x=525*tan29/[tan38-tan29]---> x=1282.12 ft
BD=x*tan 38-----> 1282.12*tan38-----> BD=1001.71 ft-----> 1002 ft
the answer is
1002 ft