Question

Please answer the questions below

1. Find the z-score that has 48.8% of the distribution's area to its left?

2. Find the z-scores for which 5% of the distribution's area lies between minus−z and z?

3. Find the indicated area under the standard normal curve. Between z=0 and z=1.34 The area between

z=0 and z=1.34 under the standard normal curve is?

Answer #2

solution

P(Z < z) = 48.8%

= P(Z < z) = 0.488

= P(Z < -0.03) = 0.488

z =-0.03 Using standard normal z table,

2.

P(-z < Z < z) = 5%=0.05

P(Z < z) - P(Z < -z) = 0.05

2 P(Z < z) - 1 = 0.05

2 P(Z < z) = 1 + 0.05 = 1.05

P(Z < z) = 1.05 / 2 = 0.525

P(Z <0.06) = 0.525

z ± 0.06 using z table

lies between -0.06 to +0.06

3.

P( 0< Z <1.34 )

= P(Z < 1.34) - P(Z < 0)

Using z table

= 0.9099-0.5

area=0.4099

z=0 and z=1.34 under the standard normal curve is 0.4099

answered by: anonymous

Find the indicated z-score. Find the indicated z-score. Find the
z-score having area 0.86 to its right under the standard normal
curve; that is, find z0.86.
Select one:
a. 0.5557
b. -1.08
c. 0.8051
d. 1.08r

Using a normal distribution and z score formula answer
the following questions
a. Find the score that cuts off the bottom 35% of the
normal curve
b. Find the data value to the nearest whole number that
cuts off the op 10% of the curve given
that the mean is 75 and sample standard deviation is
5
C Find the z scores that cut off the middle 60% of the
normal curve.

Find the z-scores for which 22% of the distribution's area
lies between minusz and z.

Find the area under the standard normal curve that lies between
the following two z values. Round your answers to at least four
decimal places.
(a)Find the area under the standard normal curve that lies
between = z − 1.28 and = z 1.36 .
(b)Find the area under the standard normal curve that lies
between = z − 2.17 and = z − 1.92 .
(c)Find the area under the standard normal curve that lies
between = z 1.36...

5.3.29 Find the z-scores for which 15% of the distribution's
area lies between minusz and z.

The values of z-score are given.
1a.) Find the area under the normal curve to the right of
z = -0.86. Round the answer to four decimal places.
The area under the normal curve to the right of z =
-0.86 is ____.
1c.) Find the area under the normal curve between z =
-0.32 and z = 0.92. Round the answer to four decimal
places.
The area under the normal curve between z = -0.32 and
z = 0.92...

Determine the area under the standard normal curve that lies to
the left of
(a) Z = -1.65 and Z = 1.65
(b) Z = -1.34 and Z=0
(c) Z = -0.15 and Z=1.81
**PLEASE include step by step so I can learn how to do this, and
please type the answer if possible, as I have a hard time following
other people's handwriting. Thanks!

1.Sketch the area under the standard normal curve over the
indicated interval and find the specified area. (Round your answer
to four decimal places.)The area to the left of
z = −1.44 is ____________
2 Sketch the area under the standard normal curve over the
indicated interval and find the specified area. (Round your answer
to four decimal places.)The area to the right of z = 1.62
is___.
3.Sketch the area under the standard normal curve over the
indicated interval...

Find the indicated z-scores shown in the graph.
Click to view page 1 of the Standard Normal Table.
LOADING...
Click to view page 2 of the Standard Normal Table.
LOADING...
z=?z=?0x0.47380.4738
A normal curve is over a horizontal x-axis and is centered on 0.
Vertical line segments extend from the curve to the horizontal axis
at two points labeled z = ? each. The area under the curve between
the left vertical line segment and 0 is shaded and labeled...

use
the standard normla table to find the z-score thag corresponds to
tbe given percentile . if the area is not in the table use the
entry closest to the area . if the area is halfway between two
entries use the z-score halfway between the corresponding z-scores
1. the z-score that corresponds to P80 is ??
(round to two decimal places as needed)
find the z -score that has 97.5% of the distribution's area to
its left
1. the...

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