Question

When investigating times required for drive-through service, the following results (in seconds) were obtained: Restaurant A 120 67 89 97 124 68 72 96 Restaurant B 115 126 49 56 98 76 78 95 Find mean, median and midrange for both restaurants. Compute the range, variance, and standard deviation for each of the two samples. Then compare the two sets of results.

Answer 1

Answer:

Step-by-step explanation:

Restaurant A data in ascending order : 67 68 72 89 96 97 120 124

Restaurant B data in ascending order : 49 56 76 78 95 98 115 126

                                    Restaurant A

• Mean         = $\frac{67+68+72+89+96+97+120+124}{8}$ =  91.625
• Median      = Since there are even number of observations so,  

                           =   $\frac{4^{th}obs + 5^{th}obs }{2}$ = $\frac{89+96}{2}$ = 92.5

• Mid range  = $\frac{Highest Value-Lowest Value}{2}$ = $\frac{124-67}{2}$ = 28.5
• Range         = Highest value - Lowest value = 124 - 67 = 57
• Variance    = $\frac{\sum (X_i - \mu )^{2}}{N-1}$ ,where Xi are sample values and $\mu$ is mean

       Solving above equation we get, variance = 493.982

• Standard Deviation = $\sqrt{variance}$ = 22.226

                                     Restaurant B

• Mean        = $\frac{49+56+76+78+95+98+115+126}{8}$ = 74.375

• Median    = Since there are even number of observations so,

                          = $\frac{4^{th}obs + 5^{th}obs }{2}$ = $\frac{78+95}{2}$ = 86.5

• Mid Range =  $\frac{Highest Value-Lowest Value}{2}$ = $\frac{126-49}{2}$ = 38.5

• Range         = Highest value - Lowest value = 126 - 49 = 77

• Variance     =  $\frac{\sum (X_i - \mu )^{2}}{N-1}$ ,where Xi are sample values and $\mu$ is mean

                            = 727.982

• Standard Deviation = $\sqrt{variance}$ = 26.981

Now comparing the results of two restaurants we conclude that:

1. Restaurant A on an average takes more time for drive-through service than Restaurant B.
2. Median time for Restaurant A is also more than Restaurant B.
3. Restaurant B has more variation in time taken for drive through service as the variance & standard deviation of Restaurant B is more than A.
4. There is more spread in the time data for Restaurant B as it has more range than A.