Find the Z-scores that separate the middle 22% of the distribution from the area in the tails of the standard normal distribution. The Z scores are ?
Find the value of z a The value of α = 0.27
The value of z 0.27 is
Find the Z-score such that the area under the standard normal curve to the left is 0.33
is the z - score such that the area under the curve to the left is 0.33.

0.279

0.613

-0.44

Step-by-step explanation:

The Zscore that seperates the middle 22% of the distribution ; could be obtained thus. The midpoint is the mean and the probability to either the right or left could be obtained.

To obtain the Zscore that seperates the middle 22% is equivalent to :

The probability to the left of the distribution :

P(x < (0.5 + 0.22/2) = p(x < 0.61)

Using the Z probability calculator :

P(x < 0.61) is equivalent to a Zscore of 0.279

α = 0.27

Z(1 - α) = Z(1 - 0.27) = Z0.73 = p(Z < 0.73) = 0.613

P(x < 0.33) = - 0.44 (Z probability calculator)

.there can be infinite values of m and n that can satisfy the above equation. for each value of m of m or n. there maybe infinite sets of solution for the above

step-by-step explanation:

step-by-step explanation:

given that there are two triangles j'k'l' and jkl>

the former is got by dilating the second one.

when drawn on a grid, (graph paper) coordinates of vertices of first triangle is

(6,12)j, (9,6)k and (12,9)l

side jk has length = distance between j and k

=

side j'k'=distance between j' and k'

=

ie j'k'l' has sides 1/3 of that of corresponding sides of jkl.

hence scale factor of 1/3 is used to get j'k'l'from jkl