Scores made on a certain aptitude test by nursing students are approximately normally distributed with a mean of 500 and a variance of 10,000. If a person is about to take the test what is the probability that he or she will make a score of 650 or more?

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Answer:

0.0668 or 6.68%

Step-by-step explanation:

Variance (V) = 10,000

Standard deviation (σ) = √V= 100

Mean score (μ) = 500

The z-score for any test score X is:

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

For X = 650:

[tex]z=\frac{650-500}{100}\\z=1.5[/tex]

A z-score of 1.5 is equivalent to the 93.32nd percentile of a normal distribution. Therefore, the probability that he or she will make a score of 650 or more is:

[tex]P(X\geq 650)=1-P(X\leq 650)\\P(X\geq 650)=1-0.9332\\P(X\geq 650)=0.0668=6.68\%[/tex]

The probability is 0.0668 or 6.68%

The probability that he or she will make a score of 650 or more is 0.0668.

Let X = Scores made on a certain aptitude test by nursing students

X follows normal distribution with mean = 500 and variance of 10,000.

So, standard deviation = [tex]\sqrt{10000}=100[/tex].

z score of 650 is = [tex]\frac{\left(650-500\right)}{100}=1.5[/tex].

The probability that he or she will make a score of 650 or more is:

[tex]P(X\geq 650)\\=P(z\geq 1.5)\\=1-P(z<1.5)\\=1-0.9332\\=0.0668[/tex]

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