Respuesta :
Answer:
The 10th percentile of the daily milk production is approximately 26.04kg.
Step-by-step explanation:
Problems of normally distributed samples can be 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 = 35, \sigma = 7[/tex]
The 10th percentile is the value of X when Z has a pvalue of 0.10. This is X when Z = -1.28. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.28 = \frac{X - 35}{7}[/tex]
[tex]X - 35 = -1.28*7[/tex]
[tex]X = 26.04[/tex]
The 10th percentile of the daily milk production is approximately 26.04kg.
Answer:
Step-by-step explanation:
Since the daily milk production of Guernsey cows is approximately normally distributed, we would apply the normal distribution formula which is expressed as
z = (x - u)/s
Where
x = daily milk production
u = mean milk production rate.
s = standard deviation
From the information given,
u = 35 kg/day
s = 7 kg/day.
The 10th percentile is 0.1. Looking at the normal distribution table, the z score corresponding to the 10th percentile is - 1.28
Therefore,
- 1.28 = (x - 35)/7
x - 35 = 7 × - 1.28
x - 35 = - 8.96
x = - 8.96 + 35
x = 26.04
The 10th percentile (in kg) of the daily milk production is approximately in 26 kg/day
