1. Find the indicated probability. a) P(0<z<1.16) b) P(_-1.42<z<0) c) P(-1.63<z<

Question

1. Find the indicated probability.
a) P(0<z<1.16)

b) P(_-1.42<z<0)

c) P(-1.63<z<1.05)

d) P(1.33<z<1.95)

e) P(10<x<18); =12, =5

f) P(x<20); =12, =5

g) P( X>9): =14, =6

h) Find z>0 such that 22% of the standard normal curves lies between 0 and z.

i) Find z<0 such that 36% of the standard normal curves lies between 0 and z.

j) Find z>0 such that28% of the standard normal curves lies between 0 and z.

2. Let x=red blood cell (RBC) count in million per cubic millimeter of whole blood.
For healthy females x has an approximately normal distribution with and convert the following to z-scores:
a) X>4.5

b) X<8.8

c) 4.2<X<6.1

Convert each of following z-interval to x-interval:
d) z <-1.44
e) z>2.1
a) z > 1.58
b) 1.45<Z<1.25

3. Average fuel consumption for Boeing 747 is 3413 gallon per hour and standard deviation 180 gallon per hour. Assume that the fuel consumption is normal distribution.
What is the probability that fuel consumption is?
a) Between 3000 and 3500 gallon per hour?

b) Less than 3000 Find gallon per hour?

amount a customer spend on lunch

4. 95% confidence interval for when =15, s =3 n=49 .

5. 98% confidence interval for when =36 , s =5, n=64 .

6. 95% confidence interval for when =15, s =4 n=15

7. 99% confidence interval for when =15, s =4 n=17 .

8. 95% confidence interval for when x=6, n=60 .

9. Consumer report gave the following information about the life (hour) size AA batteries in toys: 2,3, 5, 7, 7, and 7,8,1
a) Find mean and Standard deviation
b) Find 95% Confidence interval for life of the batteries.

10. KRLD conducted a poll of 2ooo adult and found 1400 regularly listen to an AM Radio. Let p represent adult population that regularly listens to AM Radio.
a) Find 95% confidence for p.
b) How large a sample is necessary to be 98% sure that a point estimate within 0.05 estimate of p .

Explanation / Answer

Solution:

Question 1

Part a

P(0<Z<1.16) = P(Z<1.16) – P(Z<0)

P(Z<0) = 0.50

P(Z<1.16) = 0.876976

P(0<Z<1.16) = P(Z<1.16) – P(Z<0)

P(0<Z<1.16) = 0.876976 – 0.50 = 0.376976

Required answer = 0.376976

Part b

P(-1.42<Z<0) = P(Z<0) – P(Z<-1.42)

P(Z<0) = 0.5

P(Z<-1.42) = 0.0778

P(-1.42<Z<0) = P(Z<0) – P(Z<-1.42)

P(-1.42<Z<0) = 0.5 – 0.0778

P(-1.42<Z<0) = 0.4222

Required answer = 0.4222

Part c

P(-1.63<z<1.05) = P(Z<1.05) – P(Z<-1.63)

P(Z<-1.63) = 0.051551

P(Z<1.05) = 0.853141

P(-1.63<z<1.05) = P(Z<1.05) – P(Z<-1.63)

P(-1.63<z<1.05) = 0.853141 – 0.051551

P(-1.63<z<1.05) = 0.80159

Required probability = 0.80159

Part d

P(1.33<z<1.95) = P(Z<1.95) – P(Z<1.33)

P(Z<1.95) = 0.974412

P(Z<1.33) = 0.908241

P(1.33<z<1.95) = P(Z<1.95) – P(Z<1.33)

P(1.33<z<1.95) = 0.974412 – 0.908241

P(1.33<z<1.95) = 0.066171

Required probability = 0.066171

Part e

P(10<x<18); mean=12, standard deviation =5

P(10<X<18) = P(X<18) – P(X<10)

Z score formula

Z = ( x – mean ) / standard deviation

For x = 18

Z = (18 – 12) / 5 = 6/5 = 1.2

P(Z<1.2) = 0.88493

For x = 10

Z = (10 – 12) / 5

Z = -2/5 = -0.4

P(Z<-0.4) = 0.344578

P(10<X<18) = P(X<18) – P(X<10)

P(10<X<18) = P(Z<1.2) – P(Z<-0.4)

P(10<X<18) = 0.88493 – 0.344578

P(10<X<18) = 0.540352

Required probability = 0.540352

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