Find z such that 87.7% of the standard normal curve lies to the left of z. (round your answer to two decimal places.)

z=

assume that x has a normal distribution with the specified mean and standard deviation. find the indicated probability. (round your answer to four decimal places.)

μ = 2.4; σ = 0.35

p(x ≥ 2) =

find z such that 6% of the standard normal curve lies to the right of z. (round your answer to two decimal places.)

z =

a) z = 1.16

b) P(x ≥ 2) = 0.8508

c) z = 1.555

Step-by-step explanation:

a) Find z such that 87.7% of the standard normal curve lies to the left of z.

Lies to the left, so it has a pvalue of 0.877. This is z = 1.16

b) Assume that x has a normal distribution with the specified mean and standard deviation. Find the indicated probability.

μ = 2.4; σ = 0.35

P(x ≥ 2) =

This is 1 subtracted by the pvalue of Z when X = 2, in which

has a pvalue of 0.1492.

So

P(x ≥ 2) = 1 - 0.1492 = 0.8508.

c) Find z such that 6% of the standard normal curve lies to the right of z.

To the right, so z has a pvalue of 1-0.06 = 0.94%. So z = 1.555.

2? if not the correct answer you might need to provide more information

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