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Reil [10]
3 years ago
6

The fares received by taxi drivers working for the City Taxi line are normally distributed with a mean of $12.50 and a standard

deviation of $3.25. Based on this information, what is the probability that a specific fare will exceed $15.00?
Business
1 answer:
Lyrx [107]3 years ago
3 0

Answer:

0.2308 or 23.08%

Explanation:

Mean (μ) = $12.50

Standard deviation (σ) = $3.25

Assuming a normal distribution, for any given fare X, the z-score is calculated as:

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

For X = $15.00, the z-score is:

z = \frac{15.00-12.50 }{3.25}\\ z=0.7692

A z-score of 0.7692 corresponds to the 77.91-th percentile of a normal distribution. Therefore, the probability that a fare exceeds $15.00 is:

P(X>\$15.00) = 1-0.7692 = 0.2308

The probability that a specific fare will exceed $15.00 is 0.2308.

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Answer:

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So while preparing the bank reconciliation statement, we added the deposit in transit and deduct the outstanding checks in the bank books of accounts    

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