Question

The probability that an individual has​ 20-20 vision is 0.19 0.19. in a class of 70 70 ​students, what is the mean and standard deviation of the number with​ 20-20 vision in the​ class? round to the nearest thousandth.

Answer 1

Formula to find mean and standard deviation is,

Means = np and standard deviation =$\sqrt{n p (1-p)}$.

Where n= sample size, p = probability.

According to the given problem, the probability that an individual has​ 20 vision is 0.19 in a class of 70 ​students. So,

p = 0.19 and n = 70.

Hence, the first step is to plug in these values in the above formula to get mean and standard deviation. Therefore,

Mean = np

= 70* 0.19

= 13.3

= 13.300 (Rounded to nearest thousandth.

Standard deviation = $\sqrt{np(1-p)}$

=$\sqrt{70*0.19*(1-0.19)}$

=$\sqrt{13.3*0.81}$

=√10.773

=3.282224855

= 3.282 (Rounded to nearest thousandth).

Answer 2

The mean and standard deviation of the number with​ 20-20 vision in the​ classes 13.3 and 3.282 (Rounded to the nearest thousandth).

What do you mean by standard deviation and mean?

The mean is the average value which can be calculated by dividing the sum of observations by the number of observations

In statistics, Standard deviation is a measure of the variation of a set of values.

σ = $\rm \sqrt{np(1-p)}$ standard deviation of the population

N = number of observations of population

μ = np =population mean

The mean is the average value which can be calculated by dividing the sum of observations by the number of observations

Means = np and standard deviation = $\rm \sqrt{np(1-p)}$

Where n= sample size, p = probability.

The probability that an individual has​ 20 visions is 0.19 in a class of 70 ​students. So,

p = 0.19 and n = 70.

Mean = np

= 70 × 0.19

= 13.3

= 13.300

Standard deviation =  $\rm \sqrt{np(1-p)}$

$\rm \sqrt{13.3(1-0.19)}\\\rm \sqrt{13.3(0.81)}$

= √10.773

= 3.282224855

= 3.282 (Rounded to nearest thousandth).

Learn more about standard deviation:

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