Respuesta :
Answer:
a) 2.46 standard deviations below the mean
b) The steer weighing 1000 pounds is more unusual.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
[tex]\mu = 1192, \sigma = 78[/tex]
a) How many standard deviations from the mean would a steer weighing 1000 pounds be?
This is Z when X = 1000.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1000 - 1192}{78}[/tex]
[tex]Z = -2.46[/tex]
[tex]Z = -2.46[/tex] is 2.46 standard deviations below the mean.
b) Which would be more unusual, a steer weighing 1000 pounds, or one weighing 1250 pounds?
The higher the absolute value of the z-score, the more unusual the measure is.
X = 1000 pounds
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1000 - 1192}{78}[/tex]
[tex]Z = -2.46[/tex]
The absolute value of -2.46 is 2.46
X = 1250 pounds
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{1250 - 1192}{78}[/tex]
[tex]Z = 0.74[/tex]
The absolute value of 0.74 is 0.74.
So the steer weighing 1000 pounds is more unusual.
