Question

There's a lot but its all one question. Do not answer if you do not expect to answer all of them.

A. Find the following probabilities. (Round your answers to four decimal places.)

(a)    p(0 < z < 1.45)


(b)    p(1.05 < z < 1.66)


(c)    p(−0.86 < z < 1.74)


(d)    p(z < −2.05)


(e)    p(−2.34 < z < −1.16)


(f)    p(z < 1.51)

B.Find c such that each of the following is true. (Round your answers to two decimal places.)

(a)    p(0 < z < c) = 0.1329
c =

(b)    p(c < z < 0) = 0.4802
c =

(c)    p(−c < z < c) = 0.4602
c =

(d)    p(z > c) = 0.6080
c =

(e)    p(z > c) = 0.0521
c =

(f)    p(z < c) = 0.1026
c =

C. A population X is normally distributed with mean 72.4 and standard deviation 8.4. For each of the following values of x, find the corresponding z-number. Round off your answers to two decimal places.

(a)    x = 90


(b)    x = 80


(c)    x = 75


(d)    x = 70


(e)    x = 60


(f)    x = 50

D. A population is normally distributed with mean 36.3 and standard deviation 3.5. Find the following probabilities. (Round your answers to four decimal places.)

(a)    p(36.3 < x < 39.8)


(b)    p(32.7 < x < 38.2)


(c)    p(x < 40.0)


(d)    p(31.8 < x < 40.8)


(e)    p(x = 37.4)


(f)    p(x > 37.4)

E. A population is normally distributed with mean 42.2 and standard deviation 5.7. Find the following probabilities. (Round your answers to four decimal places.)

(a)    p(42.2 < x < 47.9)


(b)    p(40.4 < x < 43.6)


(c)    p(x < 50.0)


(d)    p(32.8 < x < 51.6)


(e)    p(x = 44.8)


(f)    p(x > 44.8)

F. A population is normally distributed with mean 17.2 and standard deviation 1.7.

(a) Find the intervals representing one, two, and three standard deviations of the mean.

one standard deviation _______
two standard deviations ________
three standard deviations _________

(b) What percent of the data lies in each of the intervals in part (a)? (Round your answers to two decimal places.)

one standard deviation ____	%
two standard deviations _____	%
three standard deviations _____	%

Answer 1

A. a) P(0 < z < 1.45) = P(z < 1.45) - P(z < 0)

= 0.9265 - 0.5000

= 0.4265

The P(z < 1.45) can be obtained from the z table by finding the area corresponding to the z = 1.4 at 0.05 probability and P(z < 0) can be obtained from the z table by finding the area corresponding to the z = 0.0 at 0.00 probability.

b) P(1.05 < z < 1.66) = P(z < 1.66) - P(z < 1.05)

= 0.9515 - 0.8531

(obtained from the z table as done in previous part)

= 0.0984

c) P(-0.86 < z < 1.74) = P(z < 1.74) - P(z < -0.86)

= 0.9591 - 0.1949

= 0.7642

d) P(z < -2.05) = 0.0202

e) P(-2.34 < z < -1.16) = P(z < -1.16) - P(z < -2.34)

= 0.1230 - 0.0096

= 0.1134

f) P(z < 1.51) = 0.9345

B) a) P(0 < z < c) = 0.1329

P(z<c) - P(z<0) = 0.1329

P(z<c) - 0.5000 = 0.1329

As P(z<0) is 0.05 as obtained from the z table

So, P(z<c) = 0.1329 + 0.5000

P(z<c) = 0.6329

Hence, c will be the value of z for which the area is 0.6329

To find this we have to find z in the z table for which area is close to or equal to 0.6329.This will be at z nearly equal to 0.34.

So, c = 0.33

b) P(c < z < 0) = 0.4802

P(z < 0) - P(z < c) = 0.4802

0.5000 - P(z < c) = 0.4802

P(z < c) = 0.5000 - 0.4802 = 0.0198

Now, the area of 0.0198 corresponds to z = -2.06

Hence, c = -2.06

c) P(-c < z < c) = 0.4602

P(z < c) - P(z< -c) = 0.4602

Now, P(z < -c) = 1 - P(z < c) as z curve is symmetric about the mean ( zero)

P(z < c) -1 + P(z < c) = 0.4602

2 P(z < c) = 1.4602

P(z < c) = 0.7301

The area of 0.7301 corresponds to z = 0.61

Hence, c = 0.61

d) P(z > c) = 0.6080

P(z < c) = 1 - 0.6080 = 0.3920

The area of 0.3920 corresponds to z = -0.27

Hence, c = -0.27

e) P(z > c) = 0.0521

P(z < c) = 1 - 0.0521 = 0.9479

The area on 0.9479 corresponds to z = 1.62

Hence, c = 1.62

f) P(z < c) = 0.1026

The area of 0.1026 corresponds to z = -1.27

Hence, c = -1.27

C) The formula for z is given by -

Here, mean = 72.4

Standard deviation = 8.4

S.no.	x
a b c d e f	90 80 75 70 60 50	2.09 0.90 0.31 -0.28 -1.48 -2.67

D) mean = 36.3

Standard deviation = 3.5

Hence,

a) P(36.3 < x < 39.8) = P(0 < z < 1)

(Converting x variable into z variate as done in previous part)

= P(z < 1) - P( z < 0 )

= 0.8413 - 0.5000

= 0.3413

b) P(32.7 < x < 38.2)

= P( -1.03 < z < 0.54)

= P( z < 0.54) - P(z < -1.03)

= 0.7053 - 0.1515

= 0.5538

c) P(x < 40)

= P( z < 1.06)

= 0.8554

d) P(31.8 < x < 40.8)

= P( -1.29 < z < 1.29)

= P(z < 1.29) - P(z < -1.29)

= 0.9015 - 0.0985

= 0.8030

e) P(x = 37.4)

= P(z = 0.31)

(As )

= 0.38

f) P (x > 37.4)

= P (z > 0.31)

= 1 - P( z < 0.31)

= 1 - 0.6217

= 0.3783

F) mean = 17.2

Standard deviation = 1.7

a)The z interval for One standard deviation =[ mean - S.D , mean + S.D]

= [0-1, 0+1]

= [ -1,1]

So, corresponding x interval = [-18.9, 18.9]

The z interval for two standard deviation

= [mean - 2*S.D, mean + 2* S.D]

= [0 -2, 0+2]

= [-2 2]

Converting z variate to x variate

The corresponding x interval = [-20.6, 20.6]

The z interval for three standard deviation = [-3, 3]

The corresponding x interval = [-22.3, 22.3]

b) By property of standard normal variate, the area within

One standard deviation limit = 68.27%

Two standard deviation limit = 95.45%

Three standard deviation limit = 99.73%