Question

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:

Answer 1

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

Answer 2

The 10th percentile of the daily milk production is approximately 26.04kg.

Explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

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:

`\mu = 35, \sigma = 7`

The 10th percentile is the value of X when Z has a pvalue of 0.10. This is X when Z = -1.28. So

`Z = (X - \mu)/(\sigma)`

`-1.28 = (X - 35)/(7)`

`X - 35 = -1.28*7`

`X = 26.04`

The 10th percentile of the daily milk production is approximately 26.04kg.