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Perez9126
11.01.2020 •
Mathematics
The daily milk production of guernsey cows is approximately normally distributed with a mean of 35 kg/day and a std. deviation of 7 kg/day. the producer is concerned when the milk production of a cow falls below the 10th percentile since the animal may be ill. the 10th percentile (in kg) of the daily milk production is approximately:
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Ответ:
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
and standard deviation
, the zscore of a measure X is given by:
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:
The 10th percentile is the value of X when Z has a pvalue of 0.10. This is X when Z = -1.28. So
The 10th percentile of the daily milk production is approximately 26.04kg.
Ответ:
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
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