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aleksandrvk [35]
3 years ago
10

a magazine reports that the average bowling score for league bowlers in the united states is 157 with a standard deviation of 12

, and that the scores are approximately normally distributed
Mathematics
1 answer:
scoundrel [369]3 years ago
4 0
Given that the mean of 15 bowlers that have been selected at random is distributed normally with mean 157 and std dev of 12
The probability that a random sample of 15 bowlers would have an average score greater than 165 will be:
mean=157
std dev,σ =12
std error=σ/√n=12/√15=3.0984
standardizing xbar to z=(xbar-μ)/(σ/√n)
P(xbar>165)=P(165-157)/3.0984
=P(z>2.582)
using normal probability tables we get:
P(z>2.582)=0.0049

Next we calculate the probability that a random sample of 150 bowlers will have an average score greater than 165.
μ=157
σ=12
std error=12/√150=0.9797=0/98
standardizing the xbar we get:
z=(165-157)/0.98
=P(z>8.165)
from normal table this will give us:
P(z>8.165)=0.00


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The distribution of weights of potato chip bags filled off a production line is unknown. However, the mean is m=13.35 OZs and th
Nikitich [7]

Answer:

a) \mu_{\bar X} =13.35

e.none of the above

b) \sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02

a.0.0200

c) z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5

P(\bar X< 13.32)= P(z

a.0.0668

d) z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5

z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5

P(13.30

c.0.6853

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

Solution to the problem

Let X the random variable that represent the weights of a population, and for this case we know the distribution for X is given by:

X \sim N(13.35,0.12)  

Where \mu=13.35 and \sigma=0.12

From the central limit theorem we know that the distribution for the sample mean \bar X is given by:

\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})

Part a

\mu_{\bar X} =13.35

Part b

\sigma_{\bar X}= \frac{0.12}{\sqrt{36}}= 0.02

Part c

We want this probability:

P(\bar X< 13.32)

For this case we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got:

z = \frac{13.32-13.35}{\frac{0.12}{\sqrt{36}}}= -1.5

P(\bar X< 13.32)= P(z

Part d

We want this probability:

P(13.30

For this case we can use the z score formula given by:

z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}

And using this formula we got:

z = \frac{13.30-13.35}{\frac{0.12}{\sqrt{36}}}= -2.5

z = \frac{13.36-13.35}{\frac{0.12}{\sqrt{36}}}= 0.5

P(13.30

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If we round the lengths to the nearest tenth we get:

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