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morpeh [17]
3 years ago
6

Determine whether the given procedure results in a binomial distribution​ (or a distribution that can be treated as​ binomial).

If the procedure is not​ binomial, identify at least one requirement that is not satisfied. The YSORT method of gender​ selection, developed by the Genetics​ & IVF​ Institute, was designed to increase the likelihood that a baby will be a boy. When 120 couples use the YSORT method and give birth to 120 ​babies, the genders of the babies are recorded.
Mathematics
1 answer:
timofeeve [1]3 years ago
3 0

Answer:

The procedure results in a binomial distribution

Step-by-step explanation:

If all 120 couples used the YSORT method, it is fair to assume that the probability of a baby being a boy or a girl are constant through all trials (120 couples). Assuming 120 randomly selected couples, there is a fixed number of independent trials. Finally, since the babies can only be a boy or a girl, the binary condition is satisfied and thus, the distribution is binomial

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The travel time on a section of a Long Island Expressway (LIE) is normally distributed with a mean of 80 seconds and a standard
user100 [1]

Answer:

The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Mean of 80 seconds and a standard deviation of 6 seconds.

This means that \mu = 80, \sigma = 6

What travel time separates the top 2.5% of the travel times from the rest?

This is the 100 - 2.5 = 97.5th percentile, which is X when Z has a p-value of 0.975, so X when Z = 1.96.

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

1.96 = \frac{X - 80}{6}

X - 80 = 6*1.96

X = 91.76

The travel time that separates the top 2.5% of the travel times from the rest is of 91.76 seconds.

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Answer:

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Step-by-step explanation:

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<em><u>3</u></em><em><u>(</u></em><em><u> </u></em><em><u>x </u></em><em><u>+</u></em><em><u> </u></em><em><u>2</u></em><em><u> </u></em><em><u>)</u></em><em><u>. </u></em><em><u>=</u></em><em><u> </u></em><em><u>5</u></em><em><u>(</u></em><em><u> </u></em><em><u>x </u></em><em><u>-</u></em><em><u> </u></em><em><u>2</u></em><em><u>)</u></em>

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