49 light bulbs last average 1100 hours the standard distribution is 70
sample size (n) = 49
SD = 70
Sample SD = 70 /
= 70/7 = 10
Mean =
= 1100
= ![]()
![]()
= 1100
(70)/ ![]()
= 1100
(70)/ 7
= 1100
(10)
= 1100
20
95% confidence 1080 - 1120
Instead of sample size being known and the confidence limits unknown the situation is reversed at the 95 % confidence level
light bulbs last average 1100 hours the standard distribution is 70
5 hours ( with 95%
confidence)
-Confidence limits
required
= x +/- 2 (SD) /
1100
5 hours
1100
5 hours = 1100
2(70) / ![]()
1100 –1100
5 = 1100 – 1100
140 /
5 = 140 / ![]()
+ 5-5 = 140 /5
= 140/5
= 28
n
= ![]()
n = 784
For a binomial distribution the standard deviation is:
Where p is the
parameter. In this case,
SD =
= 0.05