The amount of calories consumed by customers at the Chinese buffet is normally distributed with mean 2984 and standard deviation
1 answer:
Answer:
a. The distribution of X is normal.
b. The probability that the customer consumes less than 2863 calories is 0.4213
c. 42.01% of of the customers consume over 3107 calories
d. The fewest number of calories a person must consume to receive the Piggy award is 4131,41 calories.
Step-by-step explanation:
Let X be the calorie of a randomly chosen customer consumes.
a. The distribution of X is normal.
b. The probability that the customer consumes less than 2863 calories =P(x<2863)=P(z<z*) where z* is the z-statistic of X=2863 calorie.
z* can be calculated using the formula
(1) z*= where
- X =2863
- M is the mean calories consumed by customers at the Chinese buffet (2984)
- s is the standard deviation (610)
Then z*= ≈−0.1984
And P(z<z*)=0.4213
c. The proportion of the customers consume over 3107 calories is
1 - the proportion of the customers consume less than 3107 calories.
= 1- P(z<z*) where z* is the z-statistic of X=3107 calorie.
From above formula (1) we get:
z*= ≈ 0.5799 and 1-0.5799=0.4201
Thus, 42.01% of of the customers consume over 3107 calories
d. The Piggy award will given out to the 3% of customers, and let z* be the z-score of the the fewest number of calories a person must consume to receive the Piggy award, then
P(z>z*)=0.03 (3%) or P(z<z*)=0.97 gives z*= 1.881
From (1), we have the equation 1.881=
We get X= (1.881×610)+2984 =4131,41
The fewest number of calories a person must consume to receive the Piggy award is 4131,41 calories.
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Answer:
a) For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
b) For this case the data is skewed to the left and we can't assume that we have the normality assumption.
c) This last case the histogram is not symmetrical and the data seems to be skewed.
Step-by-step explanation:
For this case we have the following data:
(a)data = c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
We can use the following R code to get the histogram
> x1<-c(7,13.2,8.1,8.2,6,9.5,9.4,8.7,9.8,10.9,8.4,7.4,8.4,10,9.7,8.6,12.4,10.7,11,9.4)
> hist(x1,main="Histogram a)")
The result is on the first figure attached.
For this case the histogram is not too skewed and we can say that is approximately symmetrical so then we can conclude that this dataset is similar to a normal distribution
(b)data = c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> x2<- c(2.5,1.8,2.6,-1.9,1.6,2.6,1.4,0.9,1.2,2.3,-1.5,1.5,2.5,2.9,-0.1)
> hist(x2,main="Histogram b)")
The result is on the first figure attached.
For this case the data is skewed to the left and we can't assume that we have the normality assumption.
(c)data = c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> x3<-c(3.3,1.7,3.3,3.3,2.4,0.5,1.1,1.7,12,14.4,12.8,11.2,10.9,11.7,11.7,11.6)
> hist(x3,main="Histogram c)")
The result is on the first figure attached.
This last case the histogram is not symmetrical and the data seems to be skewed.
