Pls answer both The shape of the distribution of the time required to get an oil change at a 20-minute oil-change facility is skewed right. However, records indicate that the mean time is21.3 minutes, and the standard deviation is 4.9 minutes. Complete parts (a) through (c) below.Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2).OD. The sample size needs to be less than 30.(b) What is the probability that a random sample of n = 40 oil changes results in a sample mean time less than 20 minutes?The probability is approximately 0.0467.(Round to four decimal places as needed.)(c) Suppose the manager agrees to pay each employee a $50 bonus if they meet a certain goal. On a typical Saturday, the oil-change facility will perform 40 oilchanges between 10 A.M. and 12 P.M. Treating this as a random sample, at what mean oil-change time would there be a 10% chance of being at or below? This willbe the goal established by the manager.There would be a 10% chance of being at or below minutes.(Round to one decimal place as needed.)HI'101√₁Vi1.(0,0) MoreX The reading speed of second grade students in a large city is approximately normal, with a mean of 92 words per minute (wpm) and a standard deviation of 10 wpm.Complete parts (a) through (f).Click here to view the standard normal distribution table (page 1).Click here to view the standard normal distribution table (page 2).(...)UD. Increasing the sample size increases the probability because o decreases as n increases.(e) A teacher instituted a new reading program at school. After 10 weeks in the program, it was found that the mean reading speed of a random sample of 20second grade students was 94.5 wpm. What might you conclude based on this result? Select the correct choice below and fill in the answer boxes within yourchoice.(Type integers or decimals rounded to four decimal places as needed.)OA. A mean reading rate of 94.5 wpm is not unusual since the probability of obtaining a result of 94.5 wpm or more is This means that we would expect amean reading rate of 94.5 or higher from a population whose mean reading rate is 92 in of every 100 random samples of size n = 20 students. The newprogram is not abundantly more effective than the old program.OB. A mean reading rate of 94.5 wpm is unusual since the probability of obtaining a result ofmean reading rate of 94.5 or higher from a population whose mean reading rate is 92 inprogram is abundantly more effective than the old program.94.5 wpm or more isThis means that we would expect aof every 100 random samples of size n = 20 students. The new

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