1. Refer to Section 4.1 Exercise 2 in the text. Does the curve represent a continuous probability distribution? If so, identify the type of distribution.a. Normal or Student t probability distribution b. Not a continuous probability distributionc. Uniform probability distribution d. Other continuous probability distribution2. Refer to Section 4.1 Exercise 4 in the text. Does the curve represent a continuous probability distribution? If so, identify the type of distribution.a. Normal or Student t probability distribution b. Not a continuous probability distributionc. Chi-square probability distribution d. Other continuous probability distribution3. Refer to Section 4.1 Exercise 6 in the text. Does the curve represent a continuous probability distribution? If so, identify the type of distribution.a. Normal or Student t probability distribution b. Not a continuous probability distributionc. Uniform probability distribution d. Other continuous probability distribution5. Fill in the blank to complete the uniform distribution that is described.A uniform distribution's density curve is defined by the horizontal line y = 0.25 starting at x = -2 and ending at x = ___.a. 2 b. 0c. 0.25 d. 46. For the following normal distribution, give the x-values of the inflection points of the curve (the points where the curve's concavity changes).x ~ N(0, 52)a. -10 and 10 b. 0 and 5c. -5 and 5 d. 0 and 107. For the following normal distribution, give the x-values of the inflection points of the curve (the points where the curve's concavity changes).x ~ N(-5, 1.252)a. -7.5 and -2.5 b. -6.25 and -3.75 c. -5 and 1.25 d. -5 and -2.58. The mean of a t distribution is ? = ___.a. df - 2 b. df c. 0 d. impossible to determine10. Find the z score corresponding to x = 75, ? = 80, and ? = 2.75.a. 5 b. -5 c. -1.82 d. 1.8211. Find the raw score (x) corresponding to z = 2.6, ? = 80, and ? = 2.75.a. 87.15 b. 72.85 c. 74.65 d. 85.3512. Determine the area under the standard normal curve greater than z = -2.41.a. 0.0219 b. 0.0080 c. 0.9920 d. 0.492013. Find the area under the standard normal curve between z = -1.31 and z = -1.01.a. 0.9389 b. 0.0611 c. 0.7487 d. 0.251314. Find P(z 2.75), where x ~ N(2, 1.252).a. 0.7257 b. 0.0030 c. 0.2666 d. 0.274316. Find P(-2.75 xc) = 0.09, where x ~ N(15, 2.72)a. 0.54 b. 1.34 c. 11.38 d. 18.6223. A pizzeria advertises that each of its Super Pizzas is made with 1 pound of mozzarella cheese on it. A consumer-rights advocate determines that the amount of mozzarella cheese on these pizzas has a normal distribution with a mean of 1.08 pounds and standard deviation of 0.05 pounds. What is the probability that a randomly selected Super Pizza has less than 1 pound of mozzarella cheese on it?a. 0.4681 b. 0.0548 c. 0.8413 d. 0.945224. Adult females in the United States have heights that are normally distributed with a mean of 64 inches and a standard deviation of 2.5 inches. How tall must a U.S. adult female be in order to be in the 85th percentile?a. 66.6 inches b. 61.4 inches c. 67.4 inches d. none of these25. The manufacturer of a certain brand of hot dogs claims that the mean fat content per hot dog is 20 grams. Suppose the standard deviation of the population of these hot dogs is 1.9 grams. A sample of these hot dogs is tested, and the mean fat content per hot dog of this sample is found to be 20.5 grams. Find the probability that the sample mean is at least 20.5 when the sample size is 35.a. 0.0598 b. 0.9402 c. 0.3962 d. approximately 027. A manufacturing process is designed to produce ball bearings with a diameter of 0.620 cm. The standard deviation of all ball bearing diameters is 0.009 cm. Every day a random sample of 40 ball bearings is selected and their diameters are measured. If the mean diameter from this sample falls outside of a given range, then the management will stop production to make adjustments. For the range 0.619 cm to 0.621 cm, find the probability that the management will stop production unnecessarily. (Hint: Find the probability that a sample's mean will have a value inside this range when the population mean actually is 0.620 cm, and then use complements to find the probability that a sample mean's value is outside that range.)a. 0.0885 b. 0.4822 c. 0.5178 d. 0.911528. Perform a continuity correction to rewrite the probability P(x ? 100) that involves a discrete random variable (x) as a probability involving a continuous random variable (x').a. P(x' > 99.5) b. P(x' > 100) c. P(x' > 100.5) d. P(x' > 99)29. Perform a continuity correction to rewrite the probability P(x 62.5) b. P(x' 61.5)32. Perform a continuity correction to rewrite the probability P(x is no more than 5) that involves a discrete random variable (x) as a probability involving a continuous random variable (x').a. P(x' > 5.5) b. P(x' 4.5)33. Let x be a random variable from a binomial distribution with n = 40 and p = 0.9. If a normal approximation is appropriate, give the distribution of x' that would be used in the approximation.a. x' ~ N(40, 0.92) b. x' ~ N(36, 3.62) c. x' ~ N(36, 1.92) d. normal approximation is not appropriate34. Let x be a random variable from a binomial distribution with n = 30 and p = 0.25. If a normal approximation is appropriate, give the distribution of x' that would be used in the approximation.a. x' ~ N(30, 0.252) b. x' ~ N(7.5, 5.6252) c. x' ~ N(7.5, 2.3722) d. normal approximation is not appropriate35. Of all vehicles sold by a certain car dealership, 35% are sports cars. From 50 randomly selected vehicle purchases, use the normal distribution to approximate the probability that less than 14 are sports cars.a. 0.1497 b. 0.1178 c. 0.1869 d. 0.091136. A video game system manufacturer includes a 2-year warranty with its PlayBox gaming system. Only 15% of all PlayBox systems were repaired or replaced under this warranty. A random sample of 600 PlayBox systems is selected. Use the normal distribution to approximate the probability that between 100 and 110 (inclusive) PlayBox systems were repaired or replaced under the warranty. ("inclusive" means that the endpoint values of 100 and 110 are included)a. 0.1021 b. 0.1054 c. 0.1153 d. 0.1292Section 4.1 Exercise 2,4,6 Section 4.1 Exercise 8 Section 4.1 Exercise 28

1. 2. area = 1 area = 1 -3 -2 -1 0 1 2 3 4 N CO 6 3. 0.50 - 4. 0.25- area = 1 2 0 2 4 6 8 10 12 5. 6. 0.2+ area = 0.5 area = 0.5 - 1 0 1 2 3 4 5 6 7 - 1 2 3 5