Respuesta :
Answer:
a) The difference is of 3.222 lbs.
b) 1.64 standard deviations.
c) Z = 1.64
d) Not significant, as the z-score of 1.64 is between -2 and 2.
Step-by-step explanation:
Normal Probability Distribution:
Problems of normal distributions can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.
The mean of all of the weights is x=2.088 lb, and the standard deviation of the weights is s=1.968 lb.
This means that [tex]\mu = 2.088, \sigma = 1.968[/tex]
a. What is the difference between the weight of 5.31 lb and the mean of the weights?
This is [tex]X - \mu = 5.31 - 2.088 = 3.222[/tex]
The difference is of 3.222 lbs.
b. How many standard deviations is that [the difference found in part (a)]?
This is the z-score. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{5.31 - 2.088}{1.968}[/tex]
[tex]Z = 1.64[/tex]
1.64 standard deviations.
c. Convert the weight of 5.31 lb to a z score.
Z = 1.64, as found above.
d. If we consider weights that convert to z scores between −2 and 2 to be neither significantly low nor significantly high, is the weight of 5.31 lb significant?
Not significant, as the z-score of 1.64 is between -2 and 2.
