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you want to know the percentage of medical equipment companies that earned revenue less than 64 million or more than 116 million dollars. if the mean revenue was 90 million dollars and the data has a standard deviation of 18 million, find the percentage. assume that the distribution is normal. round your answer to the nearest hundredth.

User Pcs
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Final answer:

To find the desired percentage, z-scores for $64 million and $116 million revenues were calculated, and the corresponding probabilities were obtained from the standard normal distribution table. The cumulative percentage of companies falling outside this revenue range was found to be approximately 14.98% after rounding.

Step-by-step explanation:

To find the percentage of medical equipment companies that earned revenue less than $64 million or more than $116 million, we first need to calculate the z-scores for these revenue amounts. Using the formula z = (X - μ) / σ, where X is the revenue amount, μ is the mean, and σ is the standard deviation, we can calculate the z-scores for $64 million and $116 million.

For $64 million: z = (64 - 90) / 18 ≈ -1.44

For $116 million: z = (116 - 90) / 18 ≈ 1.44

Next, we use the standard normal distribution table to find the probabilities corresponding to these z-scores. Then, we add the probability of earning less than $64 million to the probability of earning more than $116 million.

The probability of a z-score being less than -1.44 is approximately 0.0749 (or 7.49%), and the probability of a z-score being more than 1.44 is also approximately 0.0749 (as the normal distribution is symmetric).

Therefore, the total percentage of companies earning less than $64 million or more than $116 million is roughly 7.49% + 7.49% = 14.98%. Rounding to the nearest hundredth, we get 14.98%.

User Doggett
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