The number of peanuts in a 16-ounce can of Nut Munchies is normally distributed with a mean of 96.3 and a standard deviation of 2.4 peanuts. The number of peanuts in a 20-ounce can of Gone Nuts is normally distributed with a mean of 112.6 and a standard deviation of 2.8 peanuts. (a.Carmen purchased a 16-ounce can of Nut Munchies and counted 100 peanuts. What is the z-score for this can of peanuts?(b.Angelo purchased a 20-ounce can of Gone Nuts and counted 116 peanuts. What is the z-score for this can of peanuts?(c.Carmen declares that purchasing her can of Nut Munchies with 100 peanuts is less likely than Angelo purchasing a can of Gone Nuts with 116 peanuts. Is Carmen’s statement correct? Use the definition of a z-score to support or refute Carmen’s claim.

Answer:To calculate for the z-score we use the formula:z=(x-μ)/σthus the answers to questions will be as follows:Step-by-step explanation:a]Carmen purchased a 16-ounce can of Nut Munchies and counted 100 peanuts. What is the z-score for this can of peanuts?x=100μ=96.3σ=2.4z=(100-96.3)/2.4z=1.542b]Angelo purchased a 20-ounce can of Gone Nuts and counted 116 peanuts. What is the z-score for this can of peanuts?x=116μ=112.6σ=2.8thusz=(116-112.6)/2.8z=1.214c]Carmen declares that purchasing her can of Nut Munchies with 100 peanuts is less likely than Angelo purchasing a can of Gone Nuts with 116 peanuts. Is Carmen’s statement correct? Use the definition of a z-score to support or refute Carmen’s claim.This is very correct because because by definition of z-score, Munchies with 100 peanuts is 1.542 away from the mean as compared to Munchies with 116 peanuts which is 1.214 standard deviations from the mean hence the higher likelihood.