Individual assignment Essay Example

  • Category:
    Business
  • Document type:
    Assignment
  • Level:
    Undergraduate
  • Page:
    1
  • Words:
    334

Statistics Assignment

Question 1

  1. P((Z>-0.8)

Probability that Z is greater than -0.8

Applying the symmetry theory p (Z<-0.8) =1-p (Z<0.8)

Therefore p (Z<-0.8) = 1-[0.8+p (0<Z<0.8)] = 1-[0.8+0.2881] =-0.0881

  1. P(-0.85<Z<0.45)

N (0.85) = 0.8023 and N (0.45) = 0.6736

Thus N (-0.85) = 1-0.8023=0.1977

Therefore Probability= 0.6736-0.1977= 0.4759

  1. Z0.2 = 0.0793

  1. x = 70
    m = 65
    s = 4

Z= x- m / s 

P (Z>70-65/4)

P (Z>1.25) = 0.89251

  1. P(Z< 60-65/4)

P (Z<-1.25) = 0.44433

  1. P(Z>55-65/4)

(Z>-10/4) = (Z> -2.5) =0.00621

P (Z<70-65/4)

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,

  1. P ( X >750)

individual assignment

(750-800/ (100/4))

-50/25 = -2

So we get;

P( Z> -2) = 0.9772

  1. P (750<X<1000)

z-score = (data point — mean) / st. dev

individual assignment 1

Z1-score= 750-800/100

Z1-score= -0.5

Calculating the second;

Z2- score = 1000-750/100

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 z-scores. That is;

0.9938-0.3085 = 0.6853

Probability = 68.53%

  1. x = 30, 33
    m = 32
    s = 1.5

Z= x- m / s 

Z= P (30-32/1.5) = P (Z>-1.333)

Z=P (33-32/1.5) = P (Z< 0.6666)

Probability = (0.7123- 0.0934) = 0.6189

  1. Mean value μ=32 and standard deviation

σ=1.5. Suppose the sample size n=9,

Therefore, we get P(X>30)

individual assignment 2

Z= (30-32/ (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.

[500-1.96 /(12/square-root(50)]< U< [500+1.96/(12/square-root(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

[120-1.96/ (15/square-rt(51)]; [120+1.96/ (15/square-root (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> 12,000-11,500/ 800)

P (Z>0.625)

Probability =0.7324

Fewer than 10,000 pages

X= 10,000

P (Z< 10,000-11,500/ 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 Z-score and ZETA models.» Stern School of Business, New York University(2000): 9-12.