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7.4 Use Normal Distributions p. 266 Normal Distribution A bell-shaped curve is called a normal curve. It is symmetric about the mean. The percentage of the area in each standard deviation is shown above. Standard Normal Distribution Mean is 0 and standard deviation is 1. The formula is used to find the z scores. (# of standard deviations away from the mean) z x X EXAMPLE 1 Find a normal probability A normal distribution has mean x and standard deviation σ. For a randomly selected x-value from the distribution, find P(x – 2σ ≤ x ≤ x). SOLUTION The probability that a randomly selected x-value lies between x – 2σ and x is the shaded area under the normal curve shown. P( x – 2σ ≤ x ≤ x ) = 0.135 + 0.34 = 0.475 EXAMPLE 2 a. Interpret normally distribute data The blood cholesterol readings for a group of women are normally distributed with a mean of 172 mg/dl and a standard deviation of 14 mg/dl. About what percent of the women have readings between 158 and 186? (Hint: Find out how far away 158 and 186 are from the mean.) a. The readings of 158 and 186 represent one standard deviation on either side of the mean, as shown below. So, 68% of the women have readings between 158 and 186. EXAMPLE 2 Interpret normally distribute data b. Readings higher than 200 are considered undesirable. About what percent of the readings are undesirable? b. A reading of 200 is two standard deviations to the right of the mean, as shown. So, the percent of readings that are undesirable is 2.35% + 0.15%, or 2.5%. for Examples 1 and 2 GUIDED PRACTICE x standard deviation σ. A normal distribution has mean and Find the indicated probability for a randomly selected xvalue from the distribution. 1. P( x ≤ x ) 2. P( x > x ) 0.5 0.5 3. P( x < x < x + 2σ ) ANSWER 0.475 4. P( x – σ < x < x ) ANSWER 0.34 GUIDED PRACTICE 5. P(x ≤ x – 3σ) ANSWER for Examples 1 and 2 6. P(x > x + σ) 0.0015 ANSWER 0.16 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. 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.