Question Solved1 Answer 7. The number of large) inclusions in cast iron follows a Poisson distribution with a mean of 2.5 per cubic millimeter. Approximate the following probabilities (a) Determine the mean and standard deviation of the number of inclusions in a cubic centimeter (cc). (b) Approximate the probability that fewer than 2600 inclusions occur in a cc. (c) Approximate the probability that more than 2400 inclusions occur in a cc. (d) Determine the mean number of inclusions per cubic millimeter such that the probability is approximately 0.9 that 500 or fewer inclusions occur in a cc.

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Transcribed Image Text: 7. The number of large) inclusions in cast iron follows a Poisson distribution with a mean of 2.5 per cubic millimeter. Approximate the following probabilities (a) Determine the mean and standard deviation of the number of inclusions in a cubic centimeter (cc). (b) Approximate the probability that fewer than 2600 inclusions occur in a cc. (c) Approximate the probability that more than 2400 inclusions occur in a cc. (d) Determine the mean number of inclusions per cubic millimeter such that the probability is approximately 0.9 that 500 or fewer inclusions occur in a cc.
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Transcribed Image Text: 7. The number of large) inclusions in cast iron follows a Poisson distribution with a mean of 2.5 per cubic millimeter. Approximate the following probabilities (a) Determine the mean and standard deviation of the number of inclusions in a cubic centimeter (cc). (b) Approximate the probability that fewer than 2600 inclusions occur in a cc. (c) Approximate the probability that more than 2400 inclusions occur in a cc. (d) Determine the mean number of inclusions per cubic millimeter such that the probability is approximately 0.9 that 500 or fewer inclusions occur in a cc.
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ANSINER:-(a) Let x be the number of inclusions in a cubic centimeterr=2.5", then "=>lambda=2.5 xx1000=2500{:[mu=lambda],[=2500],[sigma=sqrt2500],[sigma^(2)=2500]:}{:[=2500],[sigma=sqrt2500=50quad" Here mean "=2500","" standard deviation "=50]:}(b) The probability that fewer than 2600 inclusions in a cubic centimeter can be calculated as{:[P(x < 2600)=P(x_(a) < 2600-0.5)],[=P((x_(a)-r)/(sigma) < (2599.5-2500)/(sqrt2500))],[=P(z < 1.99)],[=0.9767]:}[Ha is adjusted no. of success is 2599.5 )(C) The probability that more than 2400 inclusions in a kel{:[P(x ... See the full answer