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

(11-x)/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