G7 [s] G8 [s] H1 [s] H2 [s] H3 [s] H4 [s] H5 [s] H6 [n] /VARINFO POSITION LABEL TYPE FORMAT MEASURE ROLE VALUELABELS MISSING ATT RIBUTES RESERVEDATTRIBUTES /OPTIONS VARORDER=VARLIST SORT=ASCENDING MAXCATS=200 /STATISTICS COUNT PERCENT MEAN STDDEV QUARTILES. Codebook [DataSet1] C:\Users\admin\Desktop\MGSM816 - Market Research_Indivitual ass ignament_42403693.sav Qnaire_no Value Standard Attributes Position Label Type Format Measurement Role Valid Values 1 2 3 4 5 6 7 8 1 Questionnaire No. Numeric
Premium Standard deviation Median
is 40 months with a standard deviation of 8 months. If the life of the shoes is normally distributed‚ how many pairs of shoes out of one million will need replacement before 36 months? 308‚500 9. If the sampled population has a mean of 48 and standard deviation 16‚ then the mean and the standard deviation for the sampling distribution x for n=16 48 and 4 10. The MPG (MILES PER GALLON) for a mid-size car is normally distributed with a mean of 32 and a standard deviation of .8. what is the probability
Premium Normal distribution Standard deviation
Research Paper Trend and variability of rainfall in Tigray‚ Northern Ethiopia: Analysis of meteorological data and farmers’ perception Accepted 11th May‚ 2013 ABSTRACT Rainfall is the most important but variable climatic parameter in the semiarid tropics. In this study‚ the trend and variability of rainfall were compared with the perception of farmers in northern Ethiopia. Daily rainfall data obtained from five meteorological stations located in different agroecological zones were used
Premium Statistical significance Statistical hypothesis testing Standard deviation
012 inches and a standard deviation of 0.018 inch. The customer will buy only those pins with lengths in the interval 1.00 ( 0.02 inch‚ that is‚ between 0.98 inches to 1.02 inches. The probability of a pin having a length between 0.98 and 1.02 is 0.6339. Therefore‚ we can say that 63.39% of the pins produced will be acceptable to the customers. This is a very low acceptance level. In order to improve the percentage accepted‚ either the mean or the standard deviation needs to be adjusted. In order
Premium Normal distribution Standard deviation
by students is miles‚ and the standard deviation of the distance commuted is miles. The empirical rule states that Figure 1 • approximately of the measurements in a bell-shaped distribution lie within standard deviation of the mean; • approximately of the measurements in a bell-shaped distribution lie within standard deviations of the mean; • approximately of the measurements in a bell-shaped distribution lie within standard deviations of the mean. • • We are asked to use
Premium Standard deviation Arithmetic mean
Using the Standard Deviation You made a number of observations about the data sets for the school activities. You used mean and median to measure the center of the data‚ and you used the interquartile range (IQR) to measure the spread. When outliers are present‚ the median and IQR are used to measure center and spread because they are unaffected by extreme values. When the data appears to be symmetric and there are no known outliers‚ the mean and standard deviation (another measure of spread)
Premium Median Standard deviation Normal distribution
squared deviations‚ (d) variance‚ and (e) standard deviation: 2‚ 2‚ 0‚ 5‚ 1‚ 4‚ 1‚ 3‚ 0‚ 0‚ 1‚ 4‚ 4‚ 0‚ 1‚ 4‚ 3‚ 4‚ 2‚ 1‚ 0 Answer: A) Mean = 2 B) Median = 2 C) Sum of squared deviations =-52 D). Variance = -2.47 E). Standard deviation = 1.57 12. For the following scores‚ find the (a) mean‚ (b) median‚ (c) sum of squared deviations‚ (d) variance‚ and (e) standard deviation: 1‚112; 1‚245; 1‚361; 1‚372; 1‚472 Answer: A) Mean = 1.361 B) Median = 1.3124 C) Sum of squared deviations = 0.07608920
Premium Standard deviation Arithmetic mean
A business wants to estimate the true mean annual income of its customers. It randomly samples 220 of its customers. The mean annual income was $61‚400 with a standard deviation of $2‚200. Find a 95% confidence interval for the true mean annual income of the business’ customers. First we find E by doing Zc(standard deviation/square root of number of trials.) Now we add and subtract that number from the mean income to find both endpoints. The Zc of 95% is 1.96 so we would do 1.96((2200)/square
Premium Normal distribution Standard deviation Sample size
c) Sum of square deviations is 56. d) Variance is 2.666 or 2.7 e) Standard deviation is 1.632 or 1.6 12. a) Mean is 1312.4 or 1312 b) Median is 1361 c) Sum of square deviations is 76092.2 d) Variance is 15218.44 e) Standard deviation is 123.363 13. a) Mean is 3.166 b) Median is 3.25 c) Sum of square deviations is 0.44738 d) Variance is 0.074 e) Standard deviation is 0.272 16. a) Governor-Mean is 43 and Standard deviation is 5.916 CEO- Mean
Premium Standard deviation
caffeinated beverages consumed by Americans each day‚ based on data from the National Sleep Foundation. Determine whether or not the data represent a probability distribution. *If the data represent a probability distribution‚ find its mean and standard deviation. * If the data don’t represent a probability distribution‚ identify the requirement(s) for a probability distribution that is not satisfied. The data does not represent a probability distribution because all the probabilities are between
Free Random variable Standard deviation