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Akimi4 [234]
3 years ago
6

To be legal for certain types of prize competitions, bowling balls must be very close to 16 lb. The weights of bowling balls fro

m a certain manufacturer are known to be normally distributed with a mean of 16 lb. and a standard deviation of .25. A ball will be rejected if it weighs less than 15.68 lb.
A. What is the probability that a ball will be rejected? Round your answer to two significant digits.
B. Describe how you would use a table of random digits (comprised of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) to simulate how long it would take to identify three defective bowling balls.
C. Use the table below to perform the simulation you described in part B, this time for three defective bowling balls. Repeat your simulation three times, starting with the first digit of the first line and proceeding left to right, starting again at the extreme left of each successive line. What is the average waiting-time based on your three simulations?Describe how you would use a table of random digits (comprised of 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9) to simulate how long it would take to identify three defective bowling balls.
77014 21414 95729 01392 37813 22931 94998 56569 30213 03469
16334 43057 03297 61609 68462 26199 98324 41436 96050 95744
98563 56003 93060 29402 76577 39814 75704 26127 42577 17458
25883 51840 45515 18925 46458 61380 79369 01710 93720 73046
16434 57044 23969 78022 67976 23279 67173 44918 91684 94775
D. What is the probability that three or more defective bowing balls are found in the first 25 examined? Do
Mathematics
1 answer:
PtichkaEL [24]3 years ago
3 0

Answer:

The probability that a manufactured ball will be rejected is 0.99, to two decimal places.

Step-by-step explanation:

The mean weight is 16 Ibs. The standard deviation of the weights of all manufactured balls from the mean weight is 25. The distribution is normal. With a standard deviation this large, it is already obvious that there is a high probability of producing or having rejected balls. The weights of the balls produced here ranges from (16-25)Ibs to (16+25)Ibs.

That's -9Ibs to 41Ibs.

The minimum permissible weight is 15.68. Any ball with a lesser weight gets kicked out!

So the distance or difference between this figure and the mean is (16-15.68) = 0.32

So the probability of obtaining less than 15.68Ibs or the probability of having balls between the lower boundary of -9Ibs and 15.68 is:

25-0.32 = 24.68

24.68 ÷ 25 = 0.9872 ~ 0.99 to two decimal places.

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From the given discrete distribution, we have that:

\mu = 2.13, \sigma^2 = 0.61, \sigma = 0.78

<h3>What are the mean, the variance and the standard deviation of a discrete distribution?</h3>
  • The mean of a discrete distribution is given by the sum of each outcome multiplied by it's respective probability.
  • The variance is given by the sum of the difference squared between each observation and the mean, divided by the number of values.
  • The standard deviation is the square root of the variance.

In this problem, the distribution is:

P(X = 1) = \frac{2}{8}

P(X = 2) = \frac{3}{8}

P(X = 3) = \frac{3}{8}

Hence, the mean is:

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The variance is:

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The standard deviation is:

\sigma = \sqrt{0.61} = 0.78

More can be learned about discrete distributions at brainly.com/question/24855677

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