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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 %