Mean male height 693 in Standard deviation 28 Mean female he

Mean male height: 69.3 in.

Standard deviation: 2.8

Mean female height: 64 in.

Standard deviation: 2.8

The normal distribution presupposes that we randomly select persons or objects for our study and the persons or objects selected are independent of each other

Suppose you were to undertake a study of the average height of College male students. In general, what percent you believe will fall 1 standard deviation from the mean and what would be the heights In general, what percent you believe will fall 2 standard deviation from the mean and what would be the heights

Suppose you were to undertake a study of the average height of College female students. In general, what percent you believe will fall 1 standard deviation from the mean and what would be the heights In general, what percent you believe will fall 2 standard deviation from the mean and what would be the heights Use the following rubric as applicable and as a guide

Solution

We have give that,

Mean male height = 69.3 in.

Standard deviation = 2.8

Mean female height = 64 in.

Standard deviation = 2.8

The normal distribution presupposes that we randomly select persons or objects for our study and the persons or objects selected are independent of each other.

Suppose you were to undertake a study of the average height of College male students.

Here we have given that normal distribution.

So when distribution is Normal we use emperical rule to find a percentage.

The Emperical rule says that,

68% of the data values lie within one standard deviation of the mean.

95% of the data values lie within two standard deviation of the mean.and

99.7% of the data values lie within three standard deviation of the mean.

What percent you believe will fall 1 standard deviation from the mean and what would be the heights.

From an emperical rule we say that 68% of the data value lie within one standard deviation of the mean.

Interval of the average height of the male will be (µ + , µ - ).

µ + = 69.3 + 2.8 = 72.1

µ - = 69.3 - 2.8 = 66.5.

Interval is (72.1 , 66.5)

What percent you believe will fall 2 standard deviation from the mean and what would be the heights?

From an emperical rule we say that 95% of the data value lie within two standard deviation of the mean.

Interval of the average height of the male will be (µ + 2 , µ - 2).

µ + 2 = 69.3 + (2*2.8) = 74.9

µ - 2 =  69.3 - (2*2.8) = 63.7.

Interval is (74.9 , 63.7)

Similarly we can calculate the interval for female.

From an emperical rule we say that 68% of the data value lie within one standard deviation of the mean.

Interval of the average height of the female will be (µ + , µ - ).

µ + = 64 + 2.8 = 66.8

µ - = 64 - 2.8 = 61.2

Interval is (66.8 , 61.2)

What percent you believe will fall 2 standard deviation from the mean and what would be the heights?

From an emperical rule we say that 95% of the data value lie within two standard deviation of the mean.

Interval of the average height of the female will be (µ + 2 , µ - 2).

µ + 2 = 64 + (2*2.8) = 69.6

µ - 2 =  64 - (2*2.8) = 58.4

Interval is (69.6 , 58.4)