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vladimir2022 [97]
3 years ago
7

A random sample of 25 statistics examinations was selected. The average score in the sample was 76 with a variance of 144. Assum

ing the scores are normally distributed, the 99% confidence interval for the population average examination score is _____.
Mathematics
1 answer:
Mazyrski [523]3 years ago
6 0

Answer:

99% confidence interval for the population average examination score is between a lower limit of 69.2872 and an upper limit of 82.7128.

Step-by-step explanation:

Confidence interval = mean +/- margin of error (E)

mean = 76

variance = 144

sd = sqrt(variance) = sqrt(144) = 12

n = 25

degree of freedom (df) = n - 1 = 25 - 1 = 24

confidence level (C) = 99% = 0.99

significance level = 1 - C = 1 - 0.99 = 0.01 = 1%

t-value corresponding to 24 df and 1% significance level is 2.797

E = t×sd/√n = 2.797×12/√25 = 6.7128

Lower limit = mean - E = 76 - 6.7128 = 69.2872

Upper limit = mean + E = 76 + 6.7128 = 82.7128

99% confidence interval is (69.2872, 82.7128)

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Answer: 4.87 x 10^9  is greater than 9.212 *10^8

Step-by-step explanation:

4.87 x 10^9  is greater than 9.212 *10^8 because it has the greater exponent.

4.87 * 10^9 =  4,870,000,000

9.212 *10^8= 921,200,000

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While visiting friends in Brookfield, Janet bought a bike lock that was marked down 20% from an original price of $8.75. If the
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Answer:

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Step-by-step explanation:

Well first you have to find how much the bike lock cost.

8.75 x 0.2

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3 0
2 years ago
Whilst shopping, the probability that Caroline buys fruit is 0.7.
Lilit [14]

Answer:

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Step-by-step explanation:

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if you would let me see the list of choices.

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         Y  =  0.25 x

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