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Presentation transcript:

Homework Check

Skill Check!!! 6.1 only 

Normal Distributions

The Normal Curve Normal Distribution: Modeled by a bell-shaped curve [normal curve] Symmetrical about the mean, . Each area determined by adding or subtracting the standard deviation, . Total area under the curve is 100%, or 1.

The Normal Curve The mean is 65, and standard deviation is 2. 59 61 63 65 67 69 71 The mean is 65, and standard deviation is 2. Use this information to fill out the x-axis.

Ex. 1 Give the area under the normal curve Ex. 1 Give the area under the normal curve represented by the shaded region.

Ex. 2 Give the area under the normal curve Ex. 2 Give the area under the normal curve represented by the shaded region.

Ex. 3 A normal distribution has a mean of 18 and a Ex. 3 A normal distribution has a mean of 18 and a standard deviation of 3. Find the probability that a randomly selected x-value from the given distribution is in the interval. a. Between 12 and 18 b. At least 21

Ex. 3 A normal distribution has a mean of 18 and a Ex. 3 A normal distribution has a mean of 18 and a standard deviation of 3. Find the probability that a randomly selected x-value from the given distribution is in the interval. YOU TRY! c. At most 12 d. Between 9 and 21

4. The heights of 3000 women at a particular college are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. a) About what percentage of college women have heights below 70 inches? 97.5% b) About how many of the college women have heights between 60 inches and 65 inches? 1425 women

If it’s not on the curve: 4. The heights of 3000 women at a particular college are normally distributed with a mean of 65 inches and a standard deviation of 2.5 inches. a) What is the probability that a woman in this college would have a height less than 71 inches? If it’s not on the curve:

Using Your Calculator: TI-36X Pro: 2nd, data Distr Normalcdf Input Mean and S.D. Input Bounds CALC TI-83/84: 2nd, VARS Normalcdf( Input data as above Enter

5. A particular leg bone for dinosaur fossils has a mean length of 5 feet with standard deviation of 3 inches. What is the probability that a leg bone is less than 62 inches? = Normalcdf (-1E99, 62, 60, 3) = 0.7475

6. The weight of chocolate bars from a particular chocolate factory has a mean of 8 ounces with standard deviation of .1 ounce. What is the percent that a randomly selected bar is between 7.85 and 8.15 ounces? = Normalcdf (7.85, 8.15, 8, .1) = 86.64%

mean of 72 and a standard deviation of 8. 7. The grades on a statistics midterm exam were normally distributed with a mean of 72 and a standard deviation of 8. What is the proportion of students received a B grade. b. What is the probability that a randomly selected student received between a 65 and 85? c. What is the percent of students that failed the exam? =Normalcdf (80, 89, 72, 8) = 0.1419 =Normalcdf (65, 85, 72, 8) = 0.7571 =Normalcdf (-1E99, 69, 72, 8) = 35.38%