Download IB Math SL CH 24C

CH 24C
The Standard Normal Distribution
(Z-Distribution)
• Apples from a grower’s crop were normally distributed with
mean of 173 grams and standard deviation 34 grams. Apples
weighing less than 130 grams were too small to sell.
• Use notation to describe the information above.
X ~ N(173, 342)
• Find the proportion of apples from this crop that were too
small to sell.
P(X<130) = 10.3%
• The top 15% of the apples are sent to Trader Joes. What is the
minimum weight for an apple to be sent to Trader Joes?
k = 208g
Warm Up
• The volume of a cool drink in a bottle filled by a machine
is normally distributed with mean 503 mL and standard
deviation 0.5 mL. 1% of the bottles are rejected because
they are under-filled, and 2% are rejected because they
are over-filled; otherwise they are kept for retail. What
range of volumes is in the bottles that are kept?
Between 502 mL and 504 mL
Application Example
• The standard normal distribution or Z-distribution has a
mean of 0 and a variance of 1, hence a standard deviation
of 1.
• Every normal X distribution can be transformed into the
standard normal distribution or Z-distribution using
𝑥−𝜇
the transformation 𝑧 =
.
𝜎
• The notation for the Z-distribution is Z ~ N(0,1).
• The z-score is the number of standard deviations a data
value is above (or below) the mean.
• Z-scores are useful when comparing two populations with
different parameters 𝜇 and 𝜎.
24C
• z-scores are useful when comparing two populations
with different 𝝁 and 𝝈.
• Before graphing calculators and computer packages, it
was impossible to calculate probabilities for a general
normal distribution. 𝑁 𝜇, 𝜎 2 .
• Instead all data was transformed using the Ztransformation, and the standard normal distribution table
was consulted for the required probability values.
24C
Significance of the Z-distribution
Example 1
• The standardized verbal scores of students entering a large
university are normally distributed with a mean of 600 and a
standard deviation of 80.
• The probability that the verbal score of a student lies between 565
and 710 is represented by the shaded area in the following diagram.
This diagram represents the standard normal curve.
• Write down the values of a and b.
565 −600
𝑎=
80
𝑎 = −0.438
710 −600
b=
80
𝑏 = 1.38
a 0
Example
b
• For some questions we MUST convert to z-scores to solve.
• We always need to convert to z-scores if we are trying to find
an unknown mean or standard deviation.
Another reason z-scores
are helpful
Example:
• The weights of baby boys follow a normal distribution
with a mean of 3.3 kg. It is known that 85% of these
babies have a weight less than 3.8 kg. Find σ.
X ~ N (3.3,  2 )
P( X  3.8)  .85
3.8  3.3 

PZ 
  .85



3.8  3.3

 1.04
0.5  1.04
.482  
Example:
• The heights of certain flowers follow a normal
distribution. It is known that 20% of these flowers have
a height less than 3 cm and 10% have a height greater
than 8 cm. Find u and σ.
3  

PZ 
  0.2
 

3u
 .8416

  2.35cm
u  4.98 cm
8  

PZ 
  0.9
 

8u
 1.2816

IB Example:
• The speeds of cars at a certain point on a straight road are
normally distributed with mean u and standard deviation σ.
15 % of the cars travelled at speeds greater than 90 km h–1 and
12 % of them at speeds less than 40 km h–1. Find u and σ.
90   
40   


PZ 
PZ 
  0.85
  0.12
 
 


90  u
40  u
 1.03643
 1.17499


  22.62 km h 1
u  66.56 km h 1
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