Individual assignment Essay Example
 Category:Business
 Document type:Assignment
 Level:Undergraduate
 Page:1
 Words:334
Statistics Assignment
Question 1

P((Z>0.8)
Probability that Z is greater than 0.8
Applying the symmetry theory p (Z<0.8) =1p (Z<0.8)
Therefore p (Z<0.8) = 1[0.8+p (0<Z<0.8)] = 1[0.8+0.2881] =0.0881

P(0.85<Z<0.45)
N (0.85) = 0.8023 and N (0.45) = 0.6736
Thus N (0.85) = 10.8023=0.1977
Therefore Probability= 0.67360.1977= 0.4759

Z0.2 = 0.0793

x = 70
m = 65
s = 4
Z= x m / s
P (Z>7065/4)
P (Z>1.25) = 0.89251

P(Z< 6065/4)
P (Z<1.25) = 0.44433

P(Z>5565/4)
(Z>10/4) = (Z> 2.5) =0.00621
P (Z<7065/4)
P (Z<1.25) = 0.89251
Probability= 0.89251 0.00621=0.8863
Question 2
Mean value μ=800 and standard deviation
σ=100. Suppose the sample size n=16,

P ( X >750)
(750800/ (100/4))
50/25 = 2
So we get;
P( Z> 2) = 0.9772

P (750<X<1000)
zscore = (data point — mean) / st. dev
Z1score= 750800/100
Z1score= 0.5
Calculating the second;
Z2 score = 1000750/100
Z2 –score= 2.5
Then we proceed to find that the probability is between 0.5 and 2.5
Probability of 0.5= 0.3085
Probability of 2.5 = 0.9938
We then proceed to subtract the two zscores. That is;
0.99380.3085 = 0.6853
Probability = 68.53%

x = 30, 33
m = 32
s = 1.5
Z= x m / s
Z= P (3032/1.5) = P (Z>1.333)
Z=P (3332/1.5) = P (Z< 0.6666)
Probability = (0.7123 0.0934) = 0.6189

Mean value μ=32 and standard deviation
σ=1.5. Suppose the sample size n=9,
Therefore, we get P(X>30)
Z= (3032/ (1.5/3))
Z= 2/ 0.5 = 4
P (Z>4) = 0.9998
Question 3
Given X =500, σ=12, n=50
Determine the 95% confidence interval estimate of population mean.
[5001.96 /(12/squareroot(50)]< U< [500+1.96/(12/squareroot(50)]
498.04/1.697<U<501.96/1.697
293.48<U<295.79
[293.48; 295.79]
X =120 and s=15. ; n= 51
[1201.96/ (15/squarert(51)]; [120+1.96/ (15/squareroot (51)]
[118.04/7.1414; 121.96/7.1414]
[16.53; 17.078]
Question 4
Mean =11,500
St. deviation =800
X= 12,000
P (Z> 12,00011,500/ 800)
P (Z>0.625)
Probability =0.7324
Fewer than 10,000 pages
X= 10,000
P (Z< 10,00011,500/ 800)
P (Z< 1,500/800)
P (Z <1.875)
Probability = 0.0314
Works Cited
Altman, Edward I. «Predicting financial distress of companies: revisiting the Zscore and ZETA models.» Stern School of Business, New York University(2000): 912.