Respuesta :
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 = [tex]\frac{67+68+72+89+96+97+120+124}{8}[/tex] = 91.625
- Median = Since there are even number of observations so,
= [tex]\frac{4^{th}obs + 5^{th}obs }{2}[/tex] = [tex]\frac{89+96}{2}[/tex] = 92.5
- Mid range = [tex]\frac{Highest Value-Lowest Value}{2}[/tex] = [tex]\frac{124-67}{2}[/tex] = 28.5
- Range = Highest value - Lowest value = 124 - 67 = 57
- Variance = [tex]\frac{\sum (X_i - \mu )^{2}}{N-1}[/tex] ,where Xi are sample values and [tex]\mu[/tex] is mean
Solving above equation we get, variance = 493.982
- Standard Deviation = [tex]\sqrt{variance}[/tex] = 22.226
Restaurant B
- Mean = [tex]\frac{49+56+76+78+95+98+115+126}{8}[/tex] = 74.375
- Median = Since there are even number of observations so,
= [tex]\frac{4^{th}obs + 5^{th}obs }{2}[/tex] = [tex]\frac{78+95}{2}[/tex] = 86.5
- Mid Range = [tex]\frac{Highest Value-Lowest Value}{2}[/tex] = [tex]\frac{126-49}{2}[/tex] = 38.5
- Range = Highest value - Lowest value = 126 - 49 = 77
- Variance = [tex]\frac{\sum (X_i - \mu )^{2}}{N-1}[/tex] ,where Xi are sample values and [tex]\mu[/tex] is mean
= 727.982
- Standard Deviation = [tex]\sqrt{variance}[/tex] = 26.981
Now comparing the results of two restaurants we conclude that:
- Restaurant A on an average takes more time for drive-through service than Restaurant B.
- Median time for Restaurant A is also more than Restaurant B.
- Restaurant B has more variation in time taken for drive through service as the variance & standard deviation of Restaurant B is more than A.
- There is more spread in the time data for Restaurant B as it has more range than A.
