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A study of more than 3,000 toco toucans found that their weights were approximately Normally distributed, with a mean
of 20 ounces and a standard deviation of 2.3 ounces. Let Y be the weight of a randomly chosen toco toucan. Which of
the following correctly interprets the value of P(Y < 16)?
Find the Z-table here.
O P(Y< 16) = 0.50. About 50% of all toco toucans weigh less than 16 ounces.
OP(Y< 16) = 0.041. About 4.1% of all toco toucans weigh less than 16 ounces.
O P(Y < 16) = 0.959. About 95.9% of all toco toucans weigh less than 16 ounces.
P(Y < 16) = 0.041. About 4.1% of the toco toucans from this group weigh less than 16 ounces.
O P(Y< 16) = 0.959. About 95.9% of the toco toucans from this group weigh less than 16 ounces.

Respuesta :

Using the normal distribution, it is found that the correct option is:

  • P(Y< 16) = 0.041. About 4.1% of all toco toucans weigh less than 16 ounces

In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.

In this problem:

  • Mean of 20 ounces, hence [tex]\mu = 20[/tex].
  • Standard deviation of 2.3 ounces, hence [tex]\sigma = 2.3[/tex]

P(Y < 16) is the p-value of Z when X = 16, hence:

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

[tex]Z = \frac{16 - 20}{2.3}[/tex]

[tex]Z = -1.74[/tex]

[tex]Z = -1.74[/tex] has a p-value of 0.041.

This is valid for the entire population, hence, the correct interpretation is:

  • P(Y< 16) = 0.041. About 4.1% of all toco toucans weigh less than 16 ounces

A similar problem is given at https://brainly.com/question/24663213

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