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Halloween Candy Halloween is Kelly's favorite holiday of the year! Based on experience, Kelly knows that on average, each house gives him about 4 candies with a standard deviation of 1.5 candies. Each year Kelly visits 35 random houses in his neighborhood. Let X be the total number of candies Kelly will receive from his visits. Last Halloween, Kelly got 122 candies. Approximate the probability that the total number of candies Kelly will receive this year is smaller than last year.


Sagot :

Using the normal distribution and the central limit theorem, it is found that there is an approximately 0% probability that the total number of candies Kelly will receive this year is smaller than last year.

Normal Probability Distribution

In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], 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.
  • By the Central Limit Theorem, for n instances of a normal variable, the mean is [tex]n\mu[/tex] while the standard deviation is [tex]s = \sigma\sqrt{n}[/tex].

In this problem:

  • Mean of 4 candies, hence [tex]\mu = 4[/tex].
  • Standard deviation of 1.5 candies, hence [tex]\sigma = 1.5[/tex].
  • She visited 35 houses, hence [tex]n = 35, \mu = 35(4) = 140, s = 1.5\sqrt{4} = 3[/tex]

The probability is the p-value of Z when X = 122, hence:

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

By the Central Limit Theorem

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

[tex]Z = \frac{122 - 140}{3}[/tex]

[tex]Z = -6[/tex]

[tex]Z = -6[/tex] has a p-value of 0.

Approximately 0% probability that the total number of candies Kelly will receive this year is smaller than last year.

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

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