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Use the normal distribution of SAT critical reading scores for which the mean is 513 and the standard deviation is 110. Assume the variable x is normally distributed. (a )What percent of the SAT verbal scores are less than 650​? (b )If 1000 SAT verbal scores are randomly​ selected, about how many would you expect to be greater than 575​? (a )Approximately nothing​% of the SAT verbal scores are less than 650.

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Answer:

a) 0.8944

b) 0

Step-by-step explanation:

We are given the following information in the question:

Mean, μ = 513

Standard Deviation, σ = 110

We are given that the distribution of SAT score is a bell shaped distribution that is a normal distribution.

Formula:

[tex]z_{score} = \displaystyle\frac{x-\mu}{\sigma}[/tex]

a) P(score less than 650)

P(x < 650)

[tex]P( x < 650) = P( z < \displaystyle\frac{650 - 513}{110}) = P(z < 1.2454)[/tex]

Calculation the value from standard normal z table, we have,  

[tex]P(x < 650) = 0.8944 = 89.44\%[/tex]

b) P(score greater than 575 for 1000 randomly selected score)

[tex]P(x > 575) = P(z > \displaystyle\frac{575-513}{\frac{110}{\sqrt{1000}}}) = P(z > 17.8237)\\\\P( z > 17.8237) = 1 - P(z \leq 17.8237)[/tex]

Calculating the value from the standard normal table we have,

[tex]1 - 1= 0\\P(x > 575) = 1[/tex]

Hence, we cannot have a score greater than 575 in a randomly selected sample of 1000 scores.

Using the normal distribution, it is found that:

a) 89.44% of the SAT verbal scores are less than 650.

b) 288 of the scores would be expected to be greater than 575.

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Normal Probability Distribution  

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

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

  • 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, which is the percentile of X.

In this problem:

  • Mean of 513, thus [tex]\mu = 513[/tex]
  • Standard deviation of 110, so [tex]\sigma = 110[/tex]

Item a:

  • The proportion is the p-value of Z when X = 650, so:

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{650 - 513}{110}[/tex]

[tex]Z = 1.25[/tex]

[tex]Z = 1.25[/tex] has a p-value of 0.8944.

0.8944 x 100% = 89.44%.

89.44% of the SAT verbal scores are less than 650.

Item b:

  • First we find the proportion, which is 1 subtracted by the p-value of Z when X = 575.

[tex]Z = \frac{X - \mu}{\sigma}[/tex]

[tex]Z = \frac{575 - 513}{110}[/tex]

[tex]Z = 0.56[/tex]

[tex]Z = 0.56[/tex] has a p-value of 0.7123.

1 - 0.712 = 0.288.

Out of 1000:

0.288 x 1000 = 288

288 of the scores would be expected to be greater than 575.

A similar problem is given at https://brainly.com/question/13720222

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