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andrew11 [14]
3 years ago
12

Please help me it’s a formative grade

Mathematics
1 answer:
sergejj [24]3 years ago
5 0

Answer:

Step-by-step explanation: 800/x=100/17

(800/x)*x=(100/17)*x       - we multiply both sides of the equation by x

800=5.88235294118*x       - we divide both sides of the equation by (5.88235294118) to get x

800/5.88235294118=x

136=x

x=136

83% of 800=664

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Using the <u>normal distribution and the central limit theorem</u>, it is found that the interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

In a normal distribution with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

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

  • It 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, which is the percentile of X.
  • By the Central Limit Theorem, the sampling distribution of sample means of size n has standard deviation s = \frac{\sigma}{\sqrt{n}}.

In this problem:

  • The mean is of \mu = 3.5.
  • The standard deviation is of \sigma = 0.5.
  • Sample of 100, hence n = 100, s = \frac{0.5}{\sqrt{100}} = 0.05

The interval that contains 95.44% of the sample means for male students is <u>between Z = -2 and Z = 2</u>, as the subtraction of their p-values is 0.9544, hence:

Z = -2:

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

By the Central Limit Theorem

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

-2 = \frac{X - 3.5}{0.05}

X - 3.5 = -0.1

X = 3.4

Z = 2:

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

2 = \frac{X - 3.5}{0.05}

X - 3.5 = 0.1

X = 3.6

The interval that contains 99.44% of the sample means for male students is (3.4, 3.6).

You can learn more about the <u>normal distribution and the central limit theorem</u> at brainly.com/question/24663213

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