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S1 The Normal Distribution
1) The random variable X has a normal distribution with mean 16 and variance 0.64. Find x, such that P (X<x) = 0.025
2) Batteries for a radio have a mean life of 160 hours and a standard deviation of 30 hours. Assuming the battery life follows a normal distribution, calculate:
(a) the proportion of batteries which have a life of between 150 hours and 180 hours.
(b) the range (symmetrical about the mean) within which 75% of the batteries lie.
Answers:
1) 14.432
2a) 37.5%
2b) 125.5 hours to 194.5 hours
Cheers
2) Batteries for a radio have a mean life of 160 hours and a standard deviation of 30 hours. Assuming the battery life follows a normal distribution, calculate:
(a) the proportion of batteries which have a life of between 150 hours and 180 hours.
(b) the range (symmetrical about the mean) within which 75% of the batteries lie.
Answers:
1) 14.432
2a) 37.5%
2b) 125.5 hours to 194.5 hours
Cheers
(1)
From statistical tables, Phi(1.96) = 0.975.
So x = mean - 1.96*sd = 16 - 1.96*0.8 = 14.432.
(2)
(a)
Phi(20/30) - Phi(-10/30)
= Phi(20/30) + Phi(10/30)
= 0.378
So the proportion is 37.8% (slightly different from the given answer - what you get depends on which statistical tables you use to find the Phis).
(b)
From statistical tables, Phi(1.15) = 0.875.
The range is
(mean - 1.15*sd) to (mean - 1.15*sd)
= 125.5 to 194.5
From statistical tables, Phi(1.96) = 0.975.
So x = mean - 1.96*sd = 16 - 1.96*0.8 = 14.432.
(2)
(a)
Phi(20/30) - Phi(-10/30)
= Phi(20/30) + Phi(10/30)
= 0.378
So the proportion is 37.8% (slightly different from the given answer - what you get depends on which statistical tables you use to find the Phis).
(b)
From statistical tables, Phi(1.15) = 0.875.
The range is
(mean - 1.15*sd) to (mean - 1.15*sd)
= 125.5 to 194.5
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