
frequency distribution
can anyone check my answers for the following please? i have completed the work and want to check its right (only 3 Qs)
1. A company has 12 faulty machines. with a 70% probaility that it will last another 2 years if serviced.
a) the probability that none of the machines will last another 2 years
i have
1x((0.7)^0)x((0.3)^12)
= 5.3x10^7
b) the probability that at least 6 machines will last another 2 years
(0.7^12) + (12 x 0.7^11 x 0.3)............................. + ( 924 x 0.7^6 x 0.3^6)
= 96.1%
c) probability that at least 1 will last 2 years
i have
1  5.3 x 10^7
= 0.99999947
2 a machine produces nails of length 11cm and sd of 0.03cm. the lengths are normally distributed
a) find the % of nails between 10.94 and 11.08 cm
(11.08  11)/0.03 = 2.67
(11  10.94)/0.03 = 2
(using the z tables)
0.4962 + 0.47772 = 0.9734,
=97.34%
b) what length are 70% of the nails greater than
(11x)/0.03 = 0.84 (from 0.3 on z table)
rearanged to give x
= 10.9748 cm
3. the lengths of nails produced by a machine are known to be normally distributed, with sd 0.8. a random sample of 5 nails had lengths 106.13 106.98 104.85 107.77 105.77. using the values find a 90% confidence interval for the mean lengths of the nails
i have
mean = 106.03
106.03 + 0.59
= 106.62 or 105.44
thanks for the help