Respuesta :
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
[tex]x=6.1+1.034*1\\x=6.134[/tex]
i.e. 70% values lie between
-6.134 to +6.134
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([tex]\mu=6.1 ,\sigma^{2}= 1.0^{2}[/tex])
The z score probability distribution is given by;
Z = [tex]\frac{X-\mu}{\sigma}[/tex] ~ 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( [tex]\frac{X-\mu}{\sigma}[/tex] < [tex]\frac{x-6.1}{1}[/tex] ) = 0.30
P(Z < [tex]x-6.1[/tex]) = 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 ] .
