Question

The lifespans of Acme tires are normally distributed with a mean of 60,000 miles and a standard deviation of 5,000 miles. Compute the probability that a randomly selected tire has a lifespan between 59,000 and 61,000 miles. P(59,000 < X < 61,000) = 0.1586 (Round to four decimal places.) Describe the center and spread of the sampling distribution of for samples of size 25. ? = 60000 (Give an Exact Answer.) ? = 1000 (Give an Exact Answer.) What is the probability that a random sample of 25 tires results in a mean lifespan between 59,000 and 61,000. P(59,000 < \bar{x} < 61,000) = 0.4207 (Round to four decimal places.) What is the probability that a random sample of 100 tires results in a mean lifespan between 59,000 and 61,000. P(59,000 < \bar{x} < 61,000) = 0.1586 (Round to four decimal places.) Use the following plot to get a better understanding for your answers: 8×10^-4 7 6 5 4 -Population Distribution -Sampling Distribution, n=25 -Sampling Distribution, n=100

          The lifespans of Acme tires are normally distributed with a mean of 60,000 miles and a standard deviation of 5,000 miles.
Compute the probability that a randomly selected tire has a lifespan between 59,000 and 61,000 miles.
P(59,000 < X < 61,000) = 0.1586 (Round to four decimal places.)
Describe the center and spread of the sampling distribution of  for samples of size 25.
? = 60000 (Give an Exact Answer.)
? = 1000 (Give an Exact Answer.)
What is the probability that a random sample of 25 tires results in a mean lifespan between 59,000 and 61,000.
P(59,000 < \bar{x} < 61,000) = 0.4207 (Round to four decimal places.)
What is the probability that a random sample of 100 tires results in a mean lifespan between 59,000 and 61,000.
P(59,000 < \bar{x} < 61,000) = 0.1586 (Round to four decimal places.)
Use the following plot to get a better understanding for your answers:
8×10^-4
7
6
5
4
-Population Distribution
-Sampling Distribution, n=25
-Sampling Distribution, n=100

The lifespans of Acme tires are normally distributed with a mean of 60,000 miles and a standard deviation of 5,000 miles.
Compute the probability that a randomly selected tire has a lifespan between 59,000 and 61,000 miles.
P(59,000 < X < 61,000) = 0.1586 (Round to four decimal places.)
Describe the center and spread of the sampling distribution of for samples of size 25.
? = 60000 (Give an Exact Answer.)
? = 1000 (Give an Exact Answer.)
What is the probability that a random sample of 25 tires results in a mean lifespan between 59,000 and 61,000.
P(59,000 < x̅ < 61,000) = 0.4207 (Round to four decimal places.)
What is the probability that a random sample of 100 tires results in a mean lifespan between 59,000 and 61,000.
P(59,000 < x̅ < 61,000) = 0.1586 (Round to four decimal places.)
Use the following plot to get a better understanding for your answers:
8×10^-4
7
6
5
4
-Population Distribution
-Sampling Distribution, n=25
-Sampling Distribution, n=100

Added by Rosario L.

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