A continuous random variable X follows a normal distribution if it has the following probability density function (p.d.f.):
The parameters of the distribution are and , where m is the mean (expectation) of the distribution and is the variance. We write
X ~ N(m,s2)
to mean that the random variable X has a normal distribution with parameters m and s2.
The standard normal distribution N ( 0 , 12)
The diagram shows the standard normal distribution
Mean, µ = 0
Standard deviation, σ= 1
The general normal distribution N(µ,σ2)
Use of tables
To use the tables for a Normal distribution with mean µ and standard deviation σ
We use Z = \(\frac{X-μ}{σ}\) and look in the tables under this value of Z
Example: The length of life (in months) of Blowdri’s hair driers is approximately Normally
distributed with mean 90 months and standard deviation 15 months, N(90, 152).
(a) Each drier is sold with a 5 year guarantee. What proportion of driers fail before the guarantee expires?
(b) The manufacturer decides to change the length of the guarantee so that no more than 1% of driers fail during the guarantee period. How long should he make the guarantee?
Answer :
(a) X is the length of life of drier ⇒ X ~ N(90, 152)
5 years = 60 months
⇒ we want P(X < 60) = area up to 60
⇒ Z= \(\frac{X-μ}{σ}\) = \(\frac{60-90}{15}\) = -2.0
so we want area to left of Z = –2
Φ(–2) = 1 – Φ(2)
= 1 – 0⋅9772 = 0⋅0228 from tables.
=> the proportion of hair driers failing during the guarantee period is 0.0288 to 4 D.P.
(b) Let the length of the guarantee be t years
⇒ we need P(X < t) = 0⋅01.
We need the value of Z such that Φ(Z) = 0⋅01
From the tables Z = –2⋅3263 to 4 D.P. from tables
(remember to look in the small table after the Normal tables)
Standardising the variable
⇒ Z = \(\frac{X-μ}{σ}\) = \(\frac{t - 90}{15}\)
⇒ \(\frac{t - 90}{15}\)=-2.3263
⇒ t = 555.1 to 3 S.F.
so the manufacturer should give a guarantee period of 55 months (4 years 7 months)
Example: The results of an examination were Normally distributed. 10% of the candidates had more than 70 marks and 20% had fewer than 35 marks.
Find the mean and standard deviation of the marks.
Answer : First we need the values from the tables
⇒ Φ(–0.8416) = 0.2, and 1 – Φ(1.2816) = 0.1
Using Z = \(\frac{X-μ}{σ}\) we have
– 0⋅8416 =\(\frac{35 - μ}{σ}\)
⇒ µ = 35 + 0.8416σ …………………(1)
and 1⋅2816 =\(\frac{70 - μ}{σ}\)
⇒ µ = 70 – 1.2816σ ……………………..(2)
(1)- (2) ⇒ 0 = -35 + 2.1232σ
⇒ σ = 16.5 and µ = 48.9 to 3 S.F.
Example: The weights of chocolate bars are normally distributed with mean 205 g and standard deviation 2.6 g. The stated weight of each bar is 200 g.
(a) Find the probability that a single bar is underweight.
(b) Four bars are chosen at random. Find the probability that fewer than two bars are underweight.
Solution:
(a) Let W be the weight of a chocolate bar, W ~ N(205 , 2.62 ).
Z = \(\frac{W-μ}{σ}\) = \(\frac{200 - 205}{2.6}\) = -1.9231
P ( W < 200)=P ( Z < -1.9231)=1-Φ ( 1.92 )=1 – 0.9726
⋅
⇒ probability of an underweight bar is 0.0274.
(b) We want the probability that 0 or 1 bars chosen from 4 are underweight.
Let U be underweight and C be correct weight.
P(1 underweight) = P(CCCU) + P(CCUC) + P(CUCC) + P(UCCC)
=4×0.0274× 0.97263=0.1008
P(0 underweight) = 0.97264= 0.7404
⇒ the probability that fewer than two bars are underweight = 0.841 to 3 S.F.