Statistics Critical Values Essay - 275 Words

Statistics: Critical Values (Statistics Project Sample) Content: QUESTION 1Mean,u=16.28Standard deviation,à Ã†â€™=0.4P(x16) is given byWecompute the Z-value for the probabilityFor x=16, z= (16-16.28)/0.4=-0.7P (z-0.7) =From the standard normal curve tables, we take the area to the left of 0.5 from the region z=0 to the region z=0.7.P (z-0.7) =0.5-P (00.7)=0.5-0.2580=0.242The probability that the average amount is between 16.20 and 16.30P (16.2016.30)We compute the Z-value for the 2 figures.For x=16.20, Z= (16.20-16.28)/0.4=-0.2For x=16.30, Z= (16.30-16.28)/0.4=0.05P (-0.20.05)From the standard normal table, we compute the area to the left side from z=-0.2 up to Z=0.This will be equal to Z=0 to Z=0.2=P (-0.20.05) =0.0793+The probability that the average amount is greater than 16.50 will computed as followsWe compute the Z-value of P(x16.50)Z-value= (16.50-16.28)/0.4=0.55P (Z0.55)From the standard normal tables, we need to take the whole right side(0.5) and less the area from Z=0 to Z=0.55P (Z0.55) =0.5-0.2088=0.2912The probability that the average amount is between 16.20 and 16.26P (16.20We need to compute the z-value for the above probabilityFor x=16.20, Z-value= (16.20-16.28)/0.4=-0.2For x=16.26, Z-value = (16.26-16.28)/0.4=-0.05P (-0.05= -P (-0.2) +P (-0.05)=- [1-P (Z-0.2)] + [1-P (Z-o.o5)]=- [1-0.0793] + [1-0.0199]= -0.9207+0.9801=0.059499.9% of the time, the amount in the boxes will be 0.999*16.28=16.26 ouncesQUESTION 2When it comes to apportioning students, you can either be a male or a female.So,the probabilityof student of any given gender being selected is 50%. The number of female students is 60 %( 0.6) while the number of male students is 40 %( 0.4)Female, Q=0.6Male, P=0.4Students(male/female) 1/2 1/2 Number 0.6 0.4 The expected value,mean=1/2*6/10 +1/2*4/10=0.3+0.2=0.5The standard deviation for the random variable p is given by calculating the variance and then getting the square root.Variance=X2ip-u2=0.62*0.5 +0.42*0.5 -0.52=0.18+0.08-0.25=0.01Standard deviation=0.1The probability that less t han 58.5% of the students in the sample are female,P(à ¡Ã‚ ¿0.585)We compute the Z-value for the above probability.Z-value= (0.585-0.5)/0.1=0.085/0.1=0.85P (à ¡Ã‚ ¿0.85) =From the standard normal table, we can easily get the above probabilityP (à ¡Ã‚ ¿0.85) =0.3023The probability that more than 65% of the students are female isP (à ¡Ã‚ ¿0.65) =We compute the Z-value of the probability that, more than 65% of the students are female.Z-value= (0.65-0.5)/0.1=1.5P (Z1.5)We can obtain the value for the above probability from the normal tables.P (Z1.5) =0.4332The probability that the percentage of between 56% and 64% are female.P (0.56à ¡Ã‚ ¿0.64).The z-values areFor x=0.56, Z-value= (0.56-0.5)/0.1=0.6Therefore, P (Z0.6) =0.2257For x=0.64, Z-value= (0.64-0.50)/0.1=1.4Therefore,P (Z1.4) =0.4192The probability that the percentage of the students in the sample are female is =0.4192-0.2257=0.1935The table below is the normal table used in solving question 1. The "0.1"s are running down, w hile the "0.01"s arerunning along.Z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.090.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.03590.1 0.0398 0.0438 0.0478 0.0517 0.0557 0.0596 0.0636 0.0675 0.0714 0.07530.2 0.0793 0.0832 0.0871 0.0910 0.0948 0.0987 0.1026 0.1064 0.1103 0.11410.3 0.1179 0.1217 0.1255 0.1293 0.1331 0.1368 0.1406 0.1443 0.1480 0.15170.4 0.1554 0.1591 0.1628 0.1664 0.1700 0.1736 0.1772 0.1808 0.1844 0.18790.5 0.1915 0.195...