Originally Posted by

**Seansgirl1** I was working on some questions and got stuck. Can you review the questions and those with the question mark and question beside it are the ones I have questions on.

1. Suppose that in the 2004, 10 percent of the voters are still undecided. Find the probability that among 5 voters questioned, exactly 2 of them are undecided.

pg190 P(x)=nCx (π)__x__(1-x) __n-x __π =.10 x=2 n=5

P(2)= 5C2(.10) 2 (1-2) 5-2 = ( )(.01)(-1)

^ ^

??What is this value and how is it determined

2.On each SAT math section there are 50 questions, each question has four possible answers, one of which is correct. For students who guess at all answers, find the standard deviation for the number of correct answers.

pg 191o2=nπ (1-π) n=50 π=1/4 =50(.25)(1-50)=12.5(-49)=-612.5=612.5

o2=612.5

**3.**Doug leads bird-watching trips every morning in March. The number of cardinals seen has a Poisson distribution with a mean of 2.0. Find the probability that on a randomly selected trip, the number of cardinals seen is 2.

pg204 P(x)=__uXe____-u __u=2.0 X=2

X! P(2) = (__2.0)__ 2__ (____e____-____2.0____)__ = __(4.)(.135)__ = __.54 __= .27

2! 2! 2!

4.The ACME Scientific Instrument Company manufactures thermometers that are supposed to give readings of 0 degrees Celsius at the freezing point of water. Tests on a large sample of these thermometers reveal that at the freezing point of water, some give readings below 0 degrees (denoted by negative numbers) and some give readings above 0 degrees (denoted by positive numbers). Assume that the mean reading is 0 degrees and the standard deviation of the readings is 1.00 degrees. Also assume that the frequency distribution of errors closely resembles the normal distribution. If 7% of the thermometers are rejected because they have readings that are too high, but all other thermometers are acceptable, find the temperature that separates the rejected thermometers from the others.

Pg 229 Z=__X-u__ u=0 o=1 pr(X.>x)=0.07 z=__X-0 __z=.5000-.07=.43

01

Appendix B1 closest to .4300 is .4292 z value is 1.47

1.47=__X-0__ = X= 0- 1.47(1)=-1.47

1

5.IQ scores of UIU professors are normally distributed with a mean of 105 and a standard deviation of 21. In a random sample of 90, approximately how many Profs will have IQs between 84 and 133?

**Pg 251 n=90 u=105 o=21 Y=random variables between 84 and 133 **

**According to excel true 0.24 ****?? Please assist me in answering this problem **

6.A study of the amount of time it takes a mechanic to rebuild the transmission for a 1998 Acura Integra shows that the mean is 8.4 hours and the standard deviation is 1.8 hours. If 40 mechanics are randomly selected, find the probability that their mean rebuild time exceeds 8.1 hours. u=8.4 o=1.8/**√40**

**Pg 282 Z= **__-u __Z= __8.1 -8.4__ = __-0.3 __= -1.05 P=0.1469 found by .5000-0.3531

o**√n** 1.8/**√40 0.285 ****??****SHOULD THIS BE^ ADDED OR SUBTRACTED**

7.Find the probability that in 200 tosses of a pair fair dice, we will obtain at least 40 sevens. Assume that it is a normal distribution.

Pg 191 o2= n π(1- π) π=40 n=200 02 =200(40)(1-40)=8000(-39)=-312000

Standard deviation is 558.57

8.Use the given degree of confidence and sample data to find the margin of error in estimating the population mean μ for the following scenario. College students' annual earnings: 99% confidence; n = 67, = $6068, s = $1000.

Pg 304

± t__s __df = 67-1=66 t=2.652 = $6068 ± 2.652 __$1000 __=324.01 ± $6068

**√n****√67**

**?? what are ****The end point of confidence intervals and how did u determine this**

9.Find the margin of error for a sample of 12, where the mean is 61.4 and the standard deviation is 6.9. Use a 95% confidence interval.

Pg 298

± z__o __61.4± 1.96__6.9 __61.4±3.90

**√n****√12**

**?? what are ****The end point of confidence intervals and how did u determine this**

10.The amounts (in ounces) of juice in eight randomly selected juice bottles are:

15.8 15.9 15.9 15.6 15.2 15.7 15.8 15.6

Construct a 98 percent confidence interval for the mean amount of juice in all such bottles.

Pg 107 =125.5/8=15.7

Amt juice oz X- (X- )2

15.8 0.1 0.01

15.9 0.2 0.04

15.9 0.2 0.04

15.6 -0.1 0.01

15.2 -0.5 0.25

15.7 0 0

15.8 0.1 0.01

__15.6__ __-0.1 ____0.01__

125.5 0.37

S=**√0.37/8-1=0.2299**

± t__s __df=8-1=7 t=2.998

**√n **

125.5± 2.998 __0.2299 __=125.5 ± 0.24

**√8**

??The end points are in oz.

11.The weekly earnings of students in one age group are normally distributed with a standard deviation of 100 dollars. A researcher wishes to estimate the mean weekly earnings of students in this age group. Find the sample size needed to assure with 98 percent confidence that the sample mean will not differ from the population mean by more than 5 dollars.

Pg 316 n=(zo/E)2

The estimate sample mean will not differ from the pop mean by more than $5.00

E=5 o= 100 level of confidence=.98 ?? What is the z value of .98 confidence level and how do I determine this?