Please help solving One year consumers spent an average of $24 on a meal at a re

Question

Please help solving One year consumers spent an average of $24 on a meal at a resturant. Assume that the amount spent on a resturant meall is normaly distributed and that the standard deviation is $4. Complete parts (a) through (c) below a What is the probabity that a randomly selected person spent more than $27? P(X > $27)=(Round to four decimal places as needed ) b. What is the probability that a randomly selected person spent between $15 and $23? P($15-X-$23): (Round to four decimal places as needed ) C. Between what two values wa the middle 95% of the amounts of cash spent tain The middle 95% ofthe amounts of cash spent will fall between X- and X S Round to the nearest cent as needed ) Enter your answer in each of the answer boxes

Explanation / Answer

solution=

given = mean = 24 and standard deviation= 14

z = (X-Mean)/SD
a) z = (27-24)/14 = + 0.214
The area under the standard normal curve right to the z value indicates the required probability.
P(X>27) = P(z > 0.214) = 0.5847

b) z1 = (15-24)/14 = - 0.64
z2 = (23-24)/ 14 = - 0.07
The area under the standard normal curve between these two z values indicates the required probability.
P(15 < X < 23) = P(- 0.64 < z < - 0.07)
= 0.4721 (area corresponding to z1) - 0.2611 (area corrresponding to z2)
= 0.2110

c) Middle 95% is represented by 0.9500 area about mean. It implies that 0.9500/2 = 0.4750 area lies on left side and 0.4750 area lies on right side of mean.
The z value corresponding to 0.4750 area is 1.96
Therefore, Required X values are Mean +/- (z*SD)
24 +/- (1.96*14)
24 +/- 27.44
Middle 95% of the amounts of cash spent will fall between 3.44 and 51.44

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