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Please HIGHLIGHT answer - no detailed explanation required. PART A (only answer question with red x) A social network proyides a variety of statistics on

Please HIGHLIGHT answer - no detailed explanation required.

PART A (only answer question with red x)

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A social network proyides a variety of statistics on its website that detail the growth and popularity of the site. On average, 2? percent of 18 to 34-year-olds check their social network profiles before getting out of bed in the morning. Suppose this percentage follows a normal distribution with a standard deviation of five percent. {a} Find the probabilityr that the percent of 18 to 34-year-olds who check the social network before getting out of bed in the morning is at least 31. (Round your answer to four decimal places.) .2119 a? {b} Find the 9'51h percentile, and express it in a sentence. {Round your answer to two decimal places.) 95% of 18 to 34-year-olds who check the social network before getting out of bed in the morning is of x Find 2 such that 64% of the standard normal curve \"-33 between 2 and 2. [Round your answer to two decimal placea] I use SALT : Many people consider their smart phone to be essential! Communication, news, Internet, entertainment, photos, and just keeping current are all conveniently possible with a smart phone. However, the battery better be charged or the phone is useless. Battery life of course depends on the frequencyr duration, and type of use. One study involving heavy use of the phones showed the mean of the battery life to be 10.75 hours with a standard deviation of 2.5 hours. Then the battery needs to be recharged. Assume the battery life between charges is normally distributed. I use SALT {3) Find the probability that with heavy use, the battery life exceeds 11 hours. [Round your answer to four decimal places.) .4502 w {b} You are planning your recharging schedule so that the probability your phone will die is no more than 5%. After how many hours should you plan to recharge your phone? {Round your answer to the nearest tenth of an hour.) 14.85 3 hours Using the probabilities determined in the previous step, P(z $ 1.77) = 0.9616 and P(z

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