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# The stats homework is just like this, its required to log on and do in a timely manner

Suppose that the random variable z has a standard normal distribution. Find each of the following z points, and use the normal table to find each z point. (Round z0.03 and –z0.03 to 3 decimal places and other answers to 2 decimal places; Use the closest value of Z when there is not an exact match; if the Zvalues are equidistant, then average the two Z values. Negative values should be indicated by a minus sign.)a.z0.25[removed]b.z0.28[removed]c.z0.03[removed]d.–z0.25[removed]e.–z0.28[removed]f.–z0.03[removed]Stanford–Binet IQ Test scores are normally distributed with a mean score of 100 and a standard deviation of 14.(b)Write the equation that gives the z score corresponding to a Stanford–Binet IQ test score.z = (x – [removed] ) / [removed](c)Find the probability that a randomly selected person has an IQ test score. (Round your answers to 4 decimal places.)1.P(x > 139)[removed]2.P(x < 75)[removed]3.P(84 < x < 116)[removed]−[removed]=[removed]4.P(-2.43 < z < 2.43)[removed](d)Suppose you take the Stanford–Binet IQ Test and receive a score of 122. What percentage of people would receive a score higher than yours? (Round your answer to 2 decimal places.)A filling process is supposed to fill jars with 16 ounces of grape jelly. Specifications state that each jar must contain between 15.98 ounces and 16.02 ounces. A jar is selected from the process every half an hour until a sample of 100 jars is obtained. When the fills of the jars are measured, it is found that = 16.0024 and s = 0.02454. Using and s as point estimates of μ and σ, estimate the probability that a randomly selected jar will have a fill, x, that is out of specification. Assume that the process is in control and that the population of all jar fills is normally distributed. (Round the z-values to 2 decimal places and final answer to 4 decimal places. Negative amounts should be indicated by a minus sign.)Using the cum. normal table P ( z [removed] ) = [removed][removed] %