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Solutions to Homework #6
1. Problem
5.30
page 403
(27 points)
(5
p)
Since the average will be approximately normal, +2 standard deviations will contain 95% of the
population. For 400 radiators, the standard deviation of the average is 0.4 / 400  0.02. (2p)
Therefore the average will lie within 0.15+20.02, i.e., 0.11 to 0.19. (3p)
2. Problem
5.38
page 405
(5
p)
(a) The average will have a normal distribution with mean 55,000 miles and standard deviation =
4500 / 8  1590.99. (3p)
(b) 51,800 is 2.011 standard deviations below the mean. The probability of being this low or lower is
0.0221. (2p)
3. Problem
5.40
page 405
(7
p)
(a) The distribution is approximately normal with mean 2.2 and standard deviation 0.1941. (3p)
(b) Z=-1.03 and the probability of being less than this is 0.1515. (2p)
(c) Fewer than 100 accidents in a year corresponds to an average of < 100/52 = 1.923. For this, Z=1.427 and the probability of being less than this is 0.0768. (2p)
4. Problem
5.44
page 406
(5
p)
(a) The mean of the total is the sum of the means, 100+250=350. The standard deviation is
2.52  2.82  3.7537. (3p)
(b) The two values given are 1.332 standard deviations from the mean. The probability between them
is 0.8171. (2p)
5. Problem
5.48
page 408
(5
p)
(a) D1=2(0.002) = 0.004 and D2=2(0.001) = 0.002. (2p)
(b) Standard deviation is given by 0.002 2  0.0012  0.0012  0.002449. So D=0.005, considerably
less than the engineer’s guess of 0.008. (3p)