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Dominik [7]
3 years ago
11

In a recent year, grade 8 Washington State public school students taking a mathematics assessment test had a mean score of 281 w

ith a standard deviation of 34.4. Possible test scores could range from 0 to 500. Assume that the scores are normally distributed.
A random sample of 60 students is drawn from this population. What is the probability that the mean test score is greater than 290? Leave answers in decimal form, not percentages. Round to the nearest Hundredth (two decimal places).
Mathematics
1 answer:
sweet-ann [11.9K]3 years ago
6 0

Answer:

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = 0.0217

Step-by-step explanation:

<u><em>Step(i):</em></u>-

Mean of the Population (μ) = 281

Standard deviation of the Population = 34.4

Let 'X' be a random variable in Normal distribution

Given X = 290

Z = \frac{x -mean}{\frac{S.D}{\sqrt{n} } } = \frac{290-281}{4.44} = 2.027

<u><em>Step(ii):-</em></u>

<em> The probability that the mean test score is greater than 290</em>

P(X⁻ > 290 ) = P( Z > 2.027)

                    = 0.5 - A ( 2.027)

                   = 0.5 - 0.4783

                   = 0.0217

The probability that the mean test score is greater than 290

P(X⁻ > 290 ) = 0.0217

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Using the binomial distribution, it is found that there is a 0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

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Binomial probability distribution

P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

C_{n,x} = \frac{n!}{x!(n-x)!}

The parameters are:

  • x is the number of successes.
  • n is the number of trials.
  • p is the probability of a success on a single trial.

In this problem:

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  • A sample of 15 fatalities is taken, hence n = 15.

The probability is:

P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15)

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P(X = 15) = C_{15,15}.(0.7)^{15}.(0.3)^{0} = 0.0047

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P(10 \leq X \leq 15) = P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) = 0.2061 + 0.2186 + 0.1700 + 0.0916 + 0.0305 + 0.0047 = 0.7215

0.7215 = 72.15% probability that between 10 and 15, inclusive, accidents involved drivers who were intoxicated.

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