Question

answer part c please

A study found that the mean amount of time cars spent in drive-throughs of a certain fast-food restaurant was 134.2 seconds. Assuming drive-through times are normally distributed with a standard deviation of 34 seconds, complete parts (a) through (d) below. Click here to view the standard normal distribution table (page 1). Click here to view the standard normal distribution table (page 2). (Round to four decimal places as needed.) (b) What is the probability that a randomly selected car will spend more than 196 seconds in the restaurant's drive-through? The probability that a randomly selected car will spend more than 196 seconds in the restaurant's drive-through is 0346 (Round to four decimal places as needed.) (c) What proportion of cars spend between 2 and 3 minutes in the restaurant's drive-through? The proportion of cars that spend between 2 and 3 minutes in the restaurant's drive-through is (Round to four decimal places as needed.)

Answer 1

0.5743

M = o z 34 sics (xis fost p (120 p (1803) 120 2,= a-M 21- O. 176 -0.42 = 0.911 5= 0.3372 134-2 ~ Normally distributed with mean & Standard deviation 120 < x < 180) X is between 2 msins (120 sec) & 3 uning Lets find Zscore for 120 & 180 for 120 - 34.2 for llo u-M lfo-134.2 34 Zz z 1.347 ~ 1.35 P (126 < X< 180)= plokk 221.35 PG 2 <1:35) 0-9115 from table Plz<-0.42) p(2<0.42) =) 0.6628 (from table 0.337? P (-0.42< < < 1.35 )= P (281-35) - P(28+0,4?) Z 0.5743 2 2 - 4 -

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