Answer :
Answer:
a) The weight representing the 68th percentile is 1152 pounds.
b) The weight representing the 91st percentile is 1204 pounds.
c) The IQR for the weights of these steers is 81 pounds.
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 = 1124, \sigma = 60[/tex]
a) What weight represents the 68th percentile?
Value of X when Z has a pvalue of 0.68. So X when Z = 0.47.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.47 = \frac{X - 1124}{60}[/tex]
[tex]X - 1124 = 60*0.47[/tex]
[tex]X = 1152[/tex]
b) What weight represents the 91st percentile?
Value of X when Z has a pvalue of 0.91. So X when Z = 1.34.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]1.34 = \frac{X - 1124}{60}[/tex]
[tex]X - 1124 = 60*1.34[/tex]
[tex]X = 1204[/tex]
c) What's the IQR of the weights of these steers?
The IQR is the 75th percentile subtracted by the 25th percentile.
75th percentile
Value of X when Z has a pvalue of 0.75. So X when Z = 0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.675 = \frac{X - 1124}{60}[/tex]
[tex]X - 1124 = 60*0.675[/tex]
[tex]X = 1164.5[/tex]
25th percentile
Value of X when Z has a pvalue of 0.75. So X when Z = -0.675.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.675 = \frac{X - 1124}{60}[/tex]
[tex]X - 1124 = 60*(-0.675)[/tex]
[tex]X = 1083.5[/tex]
IQR
1164.5 - 1083.5 = 81