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Dr. Smith, a biology professor at Bradford University, has decided to give his classes a standardized biology exam that is nationally normed. This indicates that the normal distribution is an appropriate approximation for the probability distribution of students’ scores on this exam. The probability distribution of students’ scores on this standardized exam can be estimated using the normal distribution shown below.

X
X
Inflection Point
Inflection Point
70
78
86
State the mean of the distribution of the biology exam scores.

State the standard deviation of the distribution of the biology exam scores.

Grading Curve Option I
Originally, Dr. Smith decides to curve his students’ exam grades as follows.

Students whose scores are at or above the percentile will receive an A.

Students whose scores are in the – percentiles will receive a B.

Students whose scores are in the – percentiles will receive a C.

Students whose scores are in the – percentiles will receive a D.

Students whose scores are below the percentile will receive an F.

Find the z-scores that correspond to the following percentiles.

percentile

percentile

percentile

percentile

Using that information, find the exam scores that correspond to the curved grading scale. Assume that the exam scores range from 0 to 100. (Round to the nearest whole number.)

A:     –

B:     –    

C:     –    

D:     –    

F:     


Sagot :

For inflection Points of

  • 70
  • 78
  • 86

The mean and standard deviation are is mathematically given as

x=78

sigma=8

What is Normal distribution?

Normal distribution is simply a symmetrical bell-shaped graph that displays the distribution of numerous random variables.

Generally, the equation for the  mean is mathematically given as

x=total sum /amount

Therefore

x=70+78+80/3

x=78

In conclusion, Inflexion point for normal distribution is given as

[tex]I=\mu+\sigma\\\\\mu+\sigma=86[/tex]

Also

[tex]\mu-\sigma=70[/tex]

Therefore

[tex]78+\d\sigma=86[/tex]

sigma=8

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