For the wheat-yield distribution of Exercise 4.3.5, find
(a) The 65th percentile
(b) The 35th percentile
Exercise 4.3.5
In an agricultural experiment, a large uniform field was planted with a single variety of wheat. The field was divided into many plots (each plot being 7 × 100 ft) and the yield (lb) of grain was measured for each plot. These plot yields followed approximately a normal distribution with mean 88 lb and standard deviation 7 lb

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Answer:

a) 90.695 lb

b) 85.305 lb

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 = 88, \sigma = 7[/tex]

(a) The 65th percentile

X when Z has a pvalue of 0.65. So X when Z = 0.385.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]0.385 = \frac{X - 88}{7}[/tex]

[tex]X - 88 = 7*0.385[/tex]

[tex]X = 90.695[/tex]

(b) The 35th percentile

X when Z has a pvalue of 0.35. So X when Z = -0.385.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]-0.385 = \frac{X - 88}{7}[/tex]

[tex]X - 88 = 7*(-0.385)[/tex]

[tex]X = 85.305[/tex]

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