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Scores on a university exam are Normally distributed with a mean of 78 and a standard deviation of 8. The professor teaching the class declares that a score of 70 or higher is required for a grade of at least "C." Using the 68-95-99.7 rule, what percent of students score below 62? Remember that it helps to sketch the curve.

Respuesta :

Answer:

The percent of students who scored below 62 is 2.3%.

Step-by-step explanation:

In statistics, the 68–95–99.7 rule, also recognized as the empirical rule, is used to represent that 68.27%, 95.45% and 99.73% of the values of a Normally distributed data lie within one, two and three standard deviations of the mean, respectively.

Then,

  • P (-1 < Z < 1) ≈ 0.6827
  • P (-2 < Z < 2) ≈ 0.9545
  • P (3 < Z < 3) ≈ 0.9973

Given:

μ = 78

σ = 8

X = 62

Compute the distance between the value of X and μ as follows:

[tex]z=\frac{x-\mu}{\sigma}=\frac{62-78}{8}=-2[/tex]

Use the relation P (-2 < Z < 2) ≈ 0.9545 to compute the value of P (Z < -2) as follows:

[tex]P(-2<Z<2)=0.9545\\P(Z<2)-P(Z<-2)=0.9545\\1-P(Z<-2)-P(Z<-2)=0.9545\\2P(Z<-2)=1-0.9545\\P(Z<-2)=\frac{0.0455}{2}\\=0.02275[/tex]

The percentage is, 0.02275 × 100 = 2.275% ≈ 2.3%

Thus, the percent of students who scored below 62 is 2.3%.

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