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The lengths of new pencils are normally distributed with mean 11 cm and standard deviation 0.095 cm. (i)Find the probability that a pencil chosen at random has a length greater than 10.9 cm. (ii)Find the probability that, in a random sample of 6 pencils, at least two have lengths less than 10.9 cm.

User Ggariepy
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Final answer:

The probability that a pencil chosen at random has a length greater than 10.9 cm is 85.31%. In a random sample of 6 pencils, the probability that at least two have lengths less than 10.9 cm is 80.80%.

Step-by-step explanation:

(i) To find the probability that a pencil chosen at random has a length greater than 10.9 cm, we need to calculate the z-score for 10.9 cm and then find the area to the right of that z-score on the standard normal distribution table.

The z-score is calculated as (10.9 - 11) / 0.095 = -1.053.

Looking up the z-score of -1.053 on the standard normal distribution table, we find that the area to the right is 0.8531. So, the probability that a pencil chosen at random has a length greater than 10.9 cm is 0.8531 or 85.31%.

(ii) To find the probability that, in a random sample of 6 pencils, at least two have lengths less than 10.9 cm, we can calculate the probability of exactly 0 or 1 pencil having a length less than 10.9 cm (which is the complement of at least two pencils having lengths less than 10.9 cm) and subtract it from 1.

Let's calculate the probabilities:

  • Probability of 0 pencils having lengths less than 10.9 cm: ((6 choose 0) x (0.8531^0) x (0.1469^6)) = 0.0342
  • Probability of 1 pencil having a length less than 10.9 cm: ((6 choose 1) x (0.8531^1) x (0.1469^5)) = 0.1578

Adding these probabilities together, the probability of at least two pencils having lengths less than 10.9 cm is 1 - (0.0342 + 0.1578) = 0.8080 or 80.80%.

User Mustafa ASAN
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