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-Dominant- [34]
3 years ago
14

A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of sta

ndard Eskimo dogs are normally distributed, what range of weights would 99.7% of the dogs have?
Approximately 26–34 pounds
Approximately 24–36 pounds
Approximately 28–32 pounds
Approximately 29–31 pounds
Mathematics
1 answer:
nalin [4]3 years ago
5 0

Answer:

Option 2 - Approximately 24–36 pounds

Step-by-step explanation:

Given : A standard American Eskimo dog has a mean weight of 30 pounds with a standard deviation of 2 pounds. Assuming the weights of standard Eskimo dogs are normally distributed.

To find : What range of weights would 99.7% of the dogs have?

Solution :

The range of 99.7% will lie between the mean ± 3 standard deviations.

We have given,

Mean weight of Eskimo dogs is \mu=30

Standard deviation of Eskimo dogs is \sigma=2

The range of weights would 99.7% of the dogs have,

R=\mu\pm3\sigma

R=30\pm3(2)

R=30\pm6

R=30+6,30-6

R=36,24

Therefore, The range is approximately, 24 - 36 pounds.

So, Option 2 is correct.

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Answer:

The probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.

Step-by-step explanation:

According to the Central limit theorem, if from an unknown population large samples of sizes <em>n</em> > 30, are selected and the sample proportion for each sample is computed then the sampling distribution of sample proportion follows a Normal distribution.

The mean of this sampling distribution of sample proportion is:

 \mu_{\hat p}=p

The standard deviation of this sampling distribution of sample proportion is:

\sigma_{\hat p}=\sqrt{\frac{p(1-p)}{n}}

Let <em>p</em> = the proportion of keypads that pass inspection at a cell phone assembly plant.

The probability that a randomly selected cell phone keypad passes the inspection is, <em>p</em> = 0.77.

A random sample of <em>n</em> = 111 keypads is analyzed.

Then the sampling distribution of \hat p is:

\hat p\sim N(p,\ \sqrt{\frac{p(1-p)}{n}})\\\Rightarrow \hat p\sim N (0.77, 0.04)

Compute the probability that the proportion of passed keypads is between 0.72 and 0.80 as follows:

P(0.72

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Thus, the probability that the proportion of passed keypads is between 0.72 and 0.80 is 0.6677.

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