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11Alexandr11 [23.1K]
3 years ago
13

Hens usually begin laying eggs when they are about 6 months old. Young hens tend to lay smaller​ eggs, often weighing less than

the desired minimum weight of 51 grams. Complete parts​ a) through​ c) below. ​a) The average weight of the eggs produced by the young hens is 50.1 ​grams, and only 25​% of their eggs exceed the desired minimum weight. If a Normal model is​ appropriate, what would the standard deviation of the egg weights​ be?
Social Studies
1 answer:
miv72 [106K]3 years ago
6 0

Answer:

          \large\boxed{\large\boxed{1.33g}}

Explanation:

Your question has one part only: <em>a) The average weight of the eggs produced by the young hens is 50.1 ​grams, and only 25​% of their eggs exceed the desired minimum weight. If a Normal model is​ appropriate, what would the standard deviation of the egg weights​ be?</em>

<em />

<h2><em>Solution</em></h2><h2><em /></h2>

You are given the <em>mean</em>, the reference value, and the <em>percent of egss that exceeds that minimum</em>.

In terms of the parameters of a normal distribution that is:

  • <em>mean</em> =<em> 50.1g</em> (μ)
  • X<em> = 51 g</em>
  • Area of the graph above X = 51 g = <em>25%</em>

Using a standard<em> normal distribution</em> table, you can find the Z-score for which the area under the curve is greater than 25%, i.e. 0.25

The tables with two decimals for the Z-score show probability 0.2514 for Z-score of 0.67 and probabilidad 0.2483 for Z-score = 0.68.

Thus, you must interpolate. Since, (0.2514 + 0.2483)/2  ≈ 0.25, your Z-score is in the middle.

That is, Z-score = (0.67 + 0.68)/2 = 0.675.

Now use the formula for Z-score and solve for the <em>standard deviation</em> (σ):

          Z-score=(x-\mu)/\sigma

          0.675=(51g-50.1g)/\sigma

          \sigma =0.9g/0.675=1.33g

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