Scores on an exam follow an approximately bell shaped distribution with a mean of 76.4 and a standard deviation of 6.1 points. approximately, what percentage of the data is between 70.3 points and 82.5 points?

68.26%

Step-by-step explanation:

This problem can be solved by two methods by utilizing empirical rule and by using normal distribution table. The empirical rule states that

1.The area within one standard deviation contains 68.26% of the data

2.The area within two standard deviation contains 95.44% of the data

3.The area within three standard deviation contains 99.73% of the data

Here μ=76.4 and σ=6.1

μ-σ=76.4-6.1=70.3

μ+σ=76.4+6.1=82.5

The area within one standard deviation contains 68.26% of the data and so the 70.3 points and 82.5 contains 68.26% of the data.

This problem can be solved by using normal distribution table

P(70.3<X<82.5)=?

P(70.3<X<82.5)=P(70.3-76.4/6.1<Z<82.5-76.4/6.1)=P(-1<z<1)=P(-1<z<0)+P(0<z<1)

Using normal distribution we get

P(70.3<X<82.5)=0.3413+0.3413=0.6826

The percentage of the data is 0.6826*100=68.26%

problem:

f(x) = - 2/3x + 5; f(5/2)

substitute 5/2 in for x:

f(5/2) = -2/3(5/2) + 5

simplify the right side:

f(5/2) = -5/3 + 5

f(5/2) = 10/3

f(5/2) = 10/3

i hope this !

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step-by-step explanation:

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