We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5

Question

We are told that the price of basket of fries is drawn from a normal distribution with mean 6 and standard deviation of 2. You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all five baskets of fries?

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Madelyn 1 month 2021-09-11T12:03:06+00:00 1 Answer 0

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    2021-09-11T12:04:56+00:00

    Answer:

    32.64% probability that you would have enough money to pay for all five baskets of fries

    Step-by-step explanation:

    To solve this question, we have to understand the normal probability distribution and the central limit theorem.

    Normal probability distribution:

    Problems of normally distributed samples are 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 = \frac{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.

    Central limit theorem:

    The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sample means with size n can be approximated to a normal distribution with mean

    In this problem, we have that:

    \mu = 6, \sigma = 2, n = 5, s = \frac{2}{\sqrt{5}} = 0.8944

    You want to get 5 baskets of fries but you only have $28 in your pocket. What is the probability that you would have enough money to pay for all five baskets of fries?

    28/5 = 5.6

    So this is the pvalue of Z when X = 5.6.

    Z = \frac{X - \mu}{\sigma}

    By the Central Limit Theorem

    Z = \frac{X - \mu}{s}

    Z = \frac{5.6 - 6}{0.8944}

    Z = -0.45

    Z = -0.45 has a pvalue of 0.3264

    32.64% probability that you would have enough money to pay for all five baskets of fries

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