7.4 Normal 7.4 Normal DistributionsDistributions

Part IIPart IIp. 264p. 264

GUIDED PRACTICE From Yesterday’s notes

A normal distribution has mean and standard deviation σ. Find the indicated probability for a randomly selected x-value from the distribution.

x

1. P( ≤ )x x

0.5ANSWER

GUIDED PRACTICE From Yesterday’s notes

2. P( > )x x

0.5ANSWER

GUIDED PRACTICE From yesterday’s notes

3. P( < < + 2σ ) x x x

0.475ANSWER

GUIDED PRACTICE From yesterday’s notes

4. P( – σ < x < ) x x

0.34ANSWER

GUIDED PRACTICE From yesterday’s notes

5. P(x ≤ – 3σ)x

0.0015ANSWER

GUIDED PRACTICE From yesterday’s notes

6. P(x > + σ)x

0.16ANSWER

VOCABULARYVOCABULARY

Z-Score – the number (z) of standard Z-Score – the number (z) of standard deviations that a data value lies deviations that a data value lies above or below the mean of the data above or below the mean of the data set. set.

The formula below can be used to transform x-values from a normal distribution with mean and standard deviation into z-values having a standard normal distribution.

X Xz

X

EXAMPLE 3 Use a z-score and the standard normal table

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.

Biology

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.

–1.6z = x – x 50 – 7314.1=

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 for Example 3

8. WHAT IF? In Example 3, find the probability that at most 90 seals were observed during a survey.

0.8849ANSWER

GUIDED PRACTICE for Example 3

9. REASONING: Explain why it makes sense that P(z < 0) = 0.5.

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.

ANSWER

EXAMPLE 4 Use a z-score and the standard normal table

Find each person’s z-score

Matt – completed course A in 59 seconds

John – completed course B in 1 minute, 31 seconds

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.

EXAMPLE 4 Use a z-score and the standard normal table

SOLUTION

MATT Find: the z-score corresponding to an x-value of 59.

0.82z = x – x 59 – 54 6.1=

JOHNFind: the z-score corresponding to an x-value of 91.

0.69z = x – x 91 – 85 8.7=

MATT = 0.7881 or 78.8 %

JOHN = 0.7580 or 75.8 %