1.Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete parts (a) through (d). a. What is the probability that Z is less than 1.56? The probability that Z is less than 1.56 is . (Round to four decimal places as needed.) 2.Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1), complete parts (a) through (d). a.P(Z>1.021.02) b.P(Z1.021.02)= (Round to four decimal places as needed.) 3.Given a normal distribution with =100 and =10, complete parts (a) through (d). a. What is the probability that X>70? The probability that X>70 is . (Round to four decimal places as needed.) 4.The annual per capita consumption of bottled water was 32.4 gallons. Assume that the per capita consumption of bottled water is approximately normally distributed with a mean of 32.4 and a standard deviation of 13 gallons. a. What is the probability that someone consumed more than 37 gallons of bottled water? b. What is the probability that someone consumed between 25 and 35 gallons of bottled water? c. What is the probability that someone consumed less than 25 gallons of bottled water? d. 90% of people consumed less than how many gallons of bottled water? a. The probability that someone consumed more than 37 gallons of bottled water is . (Round to four decimal places as needed.) 5.A trucking company determined that the distance traveled per truck per year is normally distributed, with a mean of 80 thousand miles and a standard deviation of 10 thousand miles. Complete parts (a) through (c) below. aWhat proportion of trucks can be expected to travel between 66 and 80 thousand miles in a year? The proportion of trucks that can be expected to travel between 66 and 80 thousand miles in a year is .(Round to four decimal places as needed.) 6.Show that for a sample of n=41, the smallest and largest Zvalues are 1.98 and +1.98 and the middle (that is, 21stst) Zvalue is 0.00. With 41 observations, the smallest of the standard normal quantile values covers an area under the normal curve of 0/42=. The corresponding Zvalue is 1.98. The largest of the standard normal quantile values covers an area under the normal curve of 0/42= The corresponding Zvalue is +1.98. The middle of the standard quantile values covers an area under the normal curve of 0/42= The corresponding Zvalue is 0.00. (Type integers or decimals rounded to four decimal places as needed.) 7.One common method for measuring the size of a company is to use its market capitalization, which is computed by multiplying the number of stock shares by the price of a share of stock. The data set below contains the market capitalization values, in dollars, for 30 companies. Complete parts (a) through (c). 51.1 62.9 45.6 25.6 132. 148.3 29.3 91.2 77.6 70.9 1 257. 57.4 312.3 40.7 29.3 106.7 25.2 23.8 175.3 6 46.5 180. 122.5 202.1 255.4 63.7 77.4 199.1 121.5 8 285.7 25.5 a. Decide whether the market capitalization of these companies appears to be approximately normally distributed by comparing data characteristics to theoretical properties. Based on the theoretical properties of the normal distribution, do the data appear to be approximately normally distributed? No, the distribution of the data is significantly skewed to the right. No, because there are a significant number of outliers in the data. Yes, the distribution of the data appears to closely resemble a normal distribution. No, the distribution of the data is significantly skewed to the left. 8.Given a normal distribution with =5050 and =55, and given you select a sample of n=100, complete parts (a) through (d). a. What is the probability that X is less than 49? P( X <49)= (Type an integer or decimal rounded to four decimal places as needed.) 9.The diameter of a brand of tennis balls is approximately normally distributed, with a mean of 2.51 inches and a standard deviation of 0.05 inch. A random sample of 12 tennis balls is selected. Complete parts (a) through (d) below. a. What is the sampling distribution of the mean? A.Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 12 will not be approximately normal. B.Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 12 will also be approximately normal. C.Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 12 cannot be found. D.Because the population diameter of tennis balls is approximately normally distributed, the sampling distribution of samples of size 12 will be the uniform distribution. 10.A report announced that the median sales price of new houses sold one year was $211,000, and the mean sales price was $273,100. Assume that the standard deviation of the prices is $80,000. Complete parts (a) through (d) below. ( If you select samples of n=2, describe the shape of the sampling distribution of X. Choose the correct answer below. A.The sampling distribution will depend on the specific sample and will not have a constant shape. B.The sampling distribution will be approximately uniform. C.The sampling distribution is skewed to the right, but less skewed to the right than the population. D.The sampling distribution will be approximately normal. 11.A random sample of 64 households was selected for a phone survey. The key question asked was, "Do you or any member of your household own a product from Company A?" Of the 64 respondents, 48 said yes and 16 said no. a. Determine the sample proportion, p, of households that own a Company A product. b. If the population proportion is 0.70, determine the standard error of the proportion. a. p= (Round to two decimal places as needed.) 12.A study reports that 36% of companies in Country A have three or more female board directors. Suppose you select a random sample of 100 respondents. Complete parts (a) through (c) below. a. What is the probability that the sample will have between 33% and 41% of companies in Country A that have three or more female board directors? The probability is . (Round to four decimal places as needed.) 13.A survey of 2,450 adults reported that 58% watch news videos. Complete parts (a) through (c) below. a Suppose that you take a sample of 100 adults. If the population proportion of adults who watch news . videos is 0.58, what is the probability that fewer than half in your sample will watch news videos? The probability is that fewer than half of the adults in the sample will watch news videos. (Round to four decimal places as needed.) Consider investing in your classmate's stock and assume that the daily change is normally distributed. Using Normal.xlsx, your classmate's mean and standard deviation, determine the probability for the daily change of this stock to have: A decrease of 0.5 point or more (X -0.5)? An increase of more than 0.5 point (X > 0.5)? A decrease of more 1 point or more (X -1)? An increase of more than 1 point (X > 1)? Use the information below to answer the above questions please The daily change of this stock is: Date Open High Low Clos e Adj Close* Volume Daily Change 23-Nov-16 74.12 74.6 73.1 9 73.39 73.39 963,300 -0.73 22-Nov-16 73.61 74.6 6 73.6 1 74.49 74.49 1,281,80 0 0.88 21-Nov-16 73.07 73.6 72.8 3 73.39 73.39 1,308,70 0 0.32 18-Nov-16 73.35 73.5 4 72.8 3 73.07 73.07 1,183,20 0 -0.28 17-Nov-16 73.06 73.7 72.5 73.44 73.44 1,197,50 0.38 7 8 0 16-Nov-16 73.45 73.6 3 72.9 6 73.26 73.26 1,110,00 0 -0.19 15-Nov-16 72.73 73.5 72.7 73.4 73.4 1,884,30 0 0.67 14-Nov-16 72.43 72.9 6 71.8 3 72.53 72.53 2,320,30 0 0.1 11-Nov-16 72.72 73.2 5 72.0 5 72.5 72.5 1,784,70 0 -0.22 10-Nov-16 74.46 74.6 3 72.4 6 72.65 72.65 2,972,40 0 -1.81 9-Nov-16 75.42 75.4 5 73.3 7 74.88 74.88 1,923,10 0 -0.54 8-Nov-16 74.75 78.3 7 74.7 5 76.96 76.96 3,390,70 0 2.21 7-Nov-16 74.37 75.0 6 74.0 6 74.95 74.95 1,283,60 0 0.58 4-Nov-16 73.94 74.1 73.5 3 73.67 73.67 1,324,90 0 -0.27 3-Nov-16 74.99 75.1 2 73.7 5 73.97 73.97 1,623,40 0 -1.02 2-Nov-16 75.27 75.8 2 74.8 6 75.09 75.09 2,281,90 0 -0.18 1-Nov-16 75.62 76.0 74.7 75.21 75.21 3,058,00 -0.41 4 2 0 31-Oct-16 74.9 75.1 5 74.6 5 75.13 75.13 2,416,10 0 0.23 28-Oct-16 74.27 74.9 8 74.1 3 74.63 74.63 2,139,60 0 0.36 27-Oct-16 74.07 74.6 6 74 74.16 74.16 1,928,30 0 0.09 26-Oct-16 74.19 74.5 3 73.9 6 74.07 74.07 1,296,10 0 -0.12 25-Oct-16 74.24 74.4 9 74.1 6 74.21 74.21 1,092,80 0 -0.03 24-Oct-16 74.54 74.8 73.8 7 74.28 74.28 1,728,70 0 -0.26 21-Oct-16 74.02 74.5 8 73.6 6 74.34 74.34 1,730,70 0 0.32 20-Oct-16 74.97 74.9 7 74.0 4 74.23 74.23 1,815,80 0 -0.74 19-Oct-16 75.37 75.4 1 74.5 7 74.93 74.93 1,309,80 0 -0.44 18-Oct-16 75.44 75.6 9 75.1 2 75.36 75.36 1,282,30 0 -0.08 17-Oct-16 75.66 75.6 7 74.9 3 74.99 74.99 2,703,40 0 -0.67 14-Oct-16 75.99 76.3 75.4 75.49 75.49 2,295,30 -0.5 2 5 0 13-Oct-16 76.06 76.0 7 75.11 75.78 75.78 1,808,40 0 -0.28 12-Oct-16 76.01 76.7 2 75.8 7 76.46 76.46 1,036,80 0 0.45 11-Oct-16 76.47 76.5 2 75.7 7 76.05 76.05 1,617,70 0 -0.42 10-Oct-16 76.5 77.0 6 76.2 3 76.47 76.47 1,657,80 0 -0.03 The descriptive statistics summary table is: Daily Change Mean -0.07969697 Standard Error 0.116567451 Median -0.18 Mode 0.32 Standard Deviation Sample Variance 0.669629024 0.44840303 Kurtosis 4.088920925 Skewness 0.774119067 Range 4.02 Minimum -1.81 Maximum 2.21 Sum -2.63 Count 33 Considering the mean, median and mode, Kellogg's daily stock change distribution would be right skewed as the median of -.18 is less than the mean of -0.080