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Anastaziya [24]
2 years ago
13

The playing time X of classical CDs has the normal distribution with mean 54 and standard

Mathematics
1 answer:
pickupchik [31]2 years ago
7 0

Using the normal distribution, it is found that the relative frequency of classical CDs with playing time X between 49 and 69 minutes is 0.8403 = 84.03%.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean \mu and standard deviation \sigma is given by:

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

  • The z-score measures how many standard deviations the measure is above or below the mean.
  • Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

\mu = 54, \sigma = 5

The relative frequency of classical CDs with playing time X between 49 and 69 minutes is the <u>p-value of Z when X = 69 subtracted by the p-value of Z when X = 49</u>, hence:

X = 69:

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

Z = \frac{69 - 54}{5}

Z = 3

Z = 3 has a p-value of 0.9987.

X = 49:

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

Z = \frac{49 - 54}{5}

Z = -1

Z = -1 has a p-value of 0.1584.

0.9987 - 0.1584 = 0.8403.

More can be learned about the normal distribution at brainly.com/question/4079902

#SPJ1

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EleoNora [17]

Solution: The given problem is a binomial distribution.

The probability that a team member receives a scholarship is \frac{8}{12} = 0.67

Therefore, the given problem follows binomial with n = 5 and p = 0.6667

Now the probability that 4 of 5 players selected are on scholarship is:

P(x=4) = \binom{5}{4} \times 0.6667^{4} (1-0.6667)^{5-4}

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Therefore, the probability that 4 of 5 players selected are on scholarship is 0.329

                 

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The number of cubic units in the volume of a sphere is equal to the number of square units in the surface area of the sphere. Wh
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Answer:

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Given that

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