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max2010maxim [7]
2 years ago
13

Penny Weights, use the weights of the post 1983 pennies to construct a 98% confidence interval estimate of the standard deviatio

n of the weights of all post-1983 pennies. S=0.0165 n=37. Find confidence interval. a. 0.01231 < Ï < 0.2111 b. 0.01239 < Ï < 0.02111 c. 0.1391 < Ï < 0.2311 d. 0.001219 < Ï < 0.002111
Mathematics
1 answer:
Ronch [10]2 years ago
7 0

Answer:

c. 0.1391 < Ï < 0.2311

Step-by-step explanation:

The formula for Confidence Interval is given as:

Mean ± z × Standard deviation/√n

Z score for 98% confidence interval = 2.326

Mean = Significance level = 100% - 98%

= 2% = 0.02

Standard deviation = S=0.0165

n= 37

Hence,

Confidence Interval =

0.02 ± 2.326 × 0.0165/√37

0.02 ± 0.0063094687

Confidence Interval

0.02 - 0.0063094687

= 0.1391

0.02 ± 0.0063094687

= 0.2311

Hence, the Confidence Interval = (0.1391, 0.2311)

= c. 0.1391 < Ï < 0.2311

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Number of stocks held = 30
Price at which each shares of Lofty Cheese Company bought = 20 1/4
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Amount of profit made per share of Lofty Cheese Company = [(101/4 - (81/4)] dollars
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                                                                                                 = 5 dollars
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3 0
2 years ago
1. Kellie and her sister Ashley are training for a marathon. Kellie ran 10 mi in 75 min. Ashley ran 15 mi in 120 min. Which stat
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(a) Kellie’s minutes-per-mile pace was faster than Ashley’s minutes-per-mile pace.

True

10/75 = 0.14

15/120 = 0.13

(b) Kellie ran 8 mph.

True

10 ÷ 1.15 = 8

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False

Speed = 7.5 mph

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8 0
2 years ago
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2 years ago
The lifespan (in days) of the common housefly is best modeled using a normal curve having mean 22 days and standard deviation 5.
Natasha_Volkova [10]

Answer:

Yes, it would be unusual.

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

If Z \leq -2 or Z \geq 2, the outcome X is considered unusual.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this question, we have that:

\mu = 22, \sigma = 5, n = 25, s = \frac{5}{\sqrt{25}} = 1

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We have to find Z when X = 19. So

Z = \frac{X - \mu}{\sigma}

By the Central Limit Theorem

Z = \frac{19 - 22}{1}

Z = -3

Z = -3 \leq -2, so yes, the sample mean being less than 19 days would be considered an unusual outcome.

7 0
1 year ago
Mirela has a rectangular piece of paper with an area of 0.35 m2. She cuts the paper into small rectangles with areas of 700 cm2.
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Answer:

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So the maximum number of small rectangles she can cut

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