Question

A large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution. Determine the range of values containing 70% of the population of x

Answer 1

Answer:

Range of values containing 70% of the population of x is [-7.1364 , 7.1364 ]

Step-by-step explanation:

We are given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution.

Since, X ~ N($\mu=6.1 ,\sigma^{2}= 1.0^{2}$)

The z score probability distribution is given by;

             Z = $\frac{X-\mu}{\sigma}$ ~ N(0,1)

As we are given that we have to find the range of values containing 70% of the population of x, which means we will use 30% are on either side of the mean, i.e.;

 P(X < x) = 0.30

 P( $\frac{X-\mu}{\sigma}$ < $\frac{x-6.1}{1}$ ) = 0.30

 P(Z < $x-6.1$) = 0.30

Now, in z normal table, the critical value of x having area less than 30% is 1.0364 . i.e;

              x - 6.1 = 1.0364

               x = 7.1364

So, the range of values containing 70% of the population of x is [-7.1364 , 7.1364 ] .

               

Answer 2

Answer:

-6.134 to +6.134

Step-by-step explanation:

given that a large population of variable x is characterized by its known mean value of 6.1 units and a standard deviation of 1.0 units and a normal distribution

X is Normal with mean  =6.1 and std dev = 1 unit

We are to determine the range of values containing 70% of the population of x

We know that normal distribution curve is bell shaped symmetrical about the mean.

So to find 70% range we can use 35% on either side of the mean

Using std normal distribution table the value of z for which probability from 0 to z is 0.35 is 1.034

Hence corresponding x value is

$x=6.1+1.034*1\\x=6.134$

i.e. 70% values lie between

-6.134 to +6.134