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7.4 Normal Distributions
Part II
p. 264
GUIDED PRACTICE
From Yesterday’s notes
A normal distribution has mean x and standard
deviation σ. Find the indicated probability for a
randomly selected x-value from the distribution.
1. P( x ≤ x )
ANSWER
0.5
GUIDED PRACTICE
2. P( x > x )
ANSWER
0.5
From Yesterday’s notes
GUIDED PRACTICE
From yesterday’s notes
3. P( x < x < x + 2σ )
ANSWER
0.475
GUIDED PRACTICE
4. P( x – σ < x < x )
ANSWER
0.34
From yesterday’s notes
GUIDED PRACTICE
5. P(x ≤ x – 3σ)
ANSWER
0.0015
From yesterday’s notes
GUIDED PRACTICE
6. P(x > x + σ)
ANSWER
0.16
From yesterday’s notes
VOCABULARY

Z-Score – the number (z) of standard deviations
that a data value lies above or below the mean
of the data set.
The formula below can be used to transform x-values from a
normal distribution with mean X and standard deviation 
into z-values having a standard normal distribution.
z
X X

EXAMPLE 3
Use a z-score and the standard normal table
Biology
Scientists conducted aerial surveys of a seal sanctuary
and recorded the number x of seals they observed
during each survey. The numbers of seals observed
were normally distributed with a mean of 73 seals and a
standard deviation of 14.1 seals. Find the probability
that at most 50 seals were observed during a survey.
EXAMPLE 3
Use a z-score and the standard normal table
SOLUTION
STEP 1 Find: the z-score corresponding to an x-value
of 50.
z = x – x = 50 – 73
14.1
–1.6
STEP 2 Use: the table to find P(x < 50)
P(z < – 1.6).
The table shows that P(z < – 1.6) = 0.0548. So,
the probability that at most 50 seals were
observed during a survey is about 0.0548.
EXAMPLE 3
Use a z-score and the standard normal table
GUIDED PRACTICE
8.
for Example 3
WHAT IF? In Example 3, find the probability that at
most 90 seals were observed during a survey.
ANSWER
0.8849
GUIDED PRACTICE
9.
for Example 3
REASONING: Explain why it makes sense that
P(z < 0) = 0.5.
ANSWER
A z-score of 0 indicates that the z-score and the
mean are the same. Therefore, the area under
the normal curve is divided into two equal parts
with the mean and the z-score being equal to 0.5.
EXAMPLE 4
Use a z-score and the standard normal table
OBSTACLE COURSE
Two different obstacle courses were set up for gym
class. The times to complete Course A are normally
distributed with a mean of 54 seconds and a standard
deviation of 6.1 seconds. The times to complete
Course B are normally distributed with a mean of 1
minute, 25 seconds and a standard deviation of 8.7
seconds.
Find each person’s z-score
Matt – completed course A in 59 seconds
John – completed course B in 1 minute, 31 seconds
EXAMPLE 4
Use a z-score and the standard normal table
SOLUTION
MATT Find: the z-score corresponding to an x-value of 59.
z = x – x = 59 – 54
6.1
0.82
MATT = 0.7881 or 78.8 %
JOHN Find: the z-score corresponding to an x-value of 91.
z = x – x = 91 – 85
8.7
0.69
JOHN = 0.7580 or 75.8 %