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Statistical Calculations

Statistical Calculations

Statistical Calculations

QUESTION ONE

Three events A, B, C have probabilities

P(A) = 0.4,             P(B) = 0.2,             P(C) = 0.6.

  1. If the events are independent, calculate P (A n B), P (B/ IJ C).
  2. If the events are independent, determine the probability that at least one occurs.
  3. If P (A U B) = 0.5, then calculate P (A I B) and check if they are still independent.

Solution

P (A ∩ B) = P (A) P (B)

= (0.4) (0.2)

= 0.08

P (B/ IJ C) = P (B ∣ C) = P (B ∩ C) P (C)

= (0.3) (0.6) (0.6)

= 0.108

 

(b)

 

P(A) = 0.4

P(B) = 0.2

P(C) = 0.6

Probability of each of the events not occurring not occurring;

 

1-P(A) = 1-0.4 = 0.6

1-P(B) = 1-0.2 = 0.8

1-P(C) = 1-0.6 = 0.4

Probability that none of the events occurring is given by:

P (none of the events occurring) = (0.6) (0.8) (0.4)

= 0.192

Probability that at least one of the event occurring is:

1-(0.192) = 0.808

 

(c)

 

P (A I B) = [P (A ∩ B)]/P(B)

= 0.08/0.2

= 0.4

 

QUESTION TWO

 

Urn A contains 4 red and 8 blue chips. Urn B contains 3 red and 10 blue chips. One chip is chosen at random from urn A and moved to urn B. After that, one chip is chosen at random from urn B. Determine the probability that the chip is blue.

 

Solution

 

Urn A; 4 red chips and 8 blue chips

Urn B; 3 red chips and 10 blue chips

 

 

P (withdrawing red chip in Urn A) = (3/12)

P (withdrawing blue chip in Urn A) = (8/12)

P (Withdrawing red chip from Urn A) = (1/11)

P (Withdrawing blue chip from Urn A) = (10/11)

 

P (adding blue chip in Urn B) = (1/14)

P (adding red chip in Urn B) = (3/14)

 

 

P (blue chip) = (3/12) (1/11) + (1/14) (3/14)

= (0.25) (0.091) + (0.071) (0.21)

= 0.02275 + 0.0149

= 0.03765

 

QUESTION THREE

 

A box contains 25 red chips and 15 blue chips. We draw ten chips without replacement and mark X the number of red chips in the selection.

  1. Determine the probability distribution f (c) with its exact range.
  2. Calculate = 2).

 

Solution

 

Red chips = 25

Blue chips = 15

Total number of Chips = 40

 

(a)

F(x) = (25/40) (x/40) + (15/40) (x/40)

= (25x/160) + 15x/160

= 40x/160

= x/4

Probability of distribution of f(c) is given by replacing x by c which becomes; c/4

 

  • F (2) = 2/4

= 0.5

QUESTION FOUR

 

In a game of chance, we roll two independent six faced dice and mark a W (win) if the sum of the two faces is equal to seven and L (lose) otherwise.

In all cases a), b), c) determine

(i) the name of the distribution of X with prnf f (c), mean and variance

  1. X is the number of W when we play 20 times.

 

Probability when a dice is tossed is 1/6

When playing 20 times, the probability becomes

= (1/6)(1/20)

= 7/120

= 0.0583

  1. h) X is the number of times we play until we get the first W

Number of plays = X

1/6(x)

=x/6

 

  1. X is the number of of times we play until we obtain three W

 

1/6 (x) = 3w

 

Therefore, x/6 = 3w

 

Meaning that X = 18W

 

A continuous random variable X has probability density function  on [—1, 1] .

  1. Determine the cumulative distribution function F (c) and the percentiles Tp for all 0 < p < 1.

 

 

F(c)   =

 

(½) – 1 = 0.5

 

 

  1. Compare the median and mean (you have to calculate, not to guess)

Mean = 0.5

Median = 0.6

  1. Calculate the variance 02 .

 

 

Variance =  =  = 0.5

QUESTION FIVE

 

Let Zl, Z2 N (0, l) be two independent random variables.

 

  1. Calculate Cov (Z1 232, Zl + 3Z2) (the covariance)

 

Solution

 

Let (Z1 232) be x, and (Z1 + 3Z2) be y

 

therefore,

Cov (Z1 232, Zl + 3Z2) becomes Cov (x y)

 

 

E[XY] = E[X] E[Y]

 

From the definition,

E[XY] = E[YX]=

= )

= {}{}

= E[Y] E[X]

= E [Z1 + 3Z2] E [Z1 232]

Covariance of (Z1 232, Zl + 3Z2) is calculated as follows:

Cov (Z1-232, Zl + 3Z2) = Cov (x, y)

Cov (X, Y) = E[(Z1-232) (Zl + 3Z2)]

= E[(Z1-232)] E [(Zl + 3Z2)]

= (E[Z1] – 232) (E[Z1] + 3Z2)

= 0

Therefore;

Cov (Z1-232, Zl + 3Z2) = 0

 

P (Z1 — 24, Zl + 3Z2) (the correlation coefficient).

 

Corr (X, Y) = {Cov (X, Y)}/ {σXσY}

= {Cov (Z1 — 24) (Zl + 3Z2)}/ σ (Z1 — 24) σ) (Zl + 3Z2)

From the above definition, Cov (Z1-232, Zl + 3Z2) = 0

Therefore, correlation coefficient is zero, meaning that there is weak correlation between the two variables.

 

QUESTION SIX

 

 

Two types of batteries have lifetimes (in days) given by independent normally distributed values X, Y with distributions

X N(14.08, 0.10) ,               Y N(21.5, 0.03) .

Evaluate the probability that three batteries of type one last longer than two of type two.

 

Solution

 

The normal distribution is given by f (x, y)

Probability of the normal distribution is p {f (x, y)}

In the above case P {X, Y} = P {N (14.08, 0.10), N (21.5, 0.03)}

 

X ~ N (14.08, 0.10), Y~ N {21.5, 0.03}

P (13.98< X < 14.18) = p[(13.98 – 14. 18)]/ 0.1< Z <[(14.18 – 14.08)]/ 0.1

P(-0.2/0.1) < Z < 0.1/0.1

P (-2< X <1)

From the Z tables,

0.0202 – 0.8289 = -0.8087

 

For Y~ N {21.5, 0.03}

 

P (21.47 < X < 21.53) = P({21.47– 21.5}/ 0.03) < (21.53 – 21.5)/ 0.03

 

P (-1 < x < 1)

From Z tables,

-0.1469 – 0.8531 = -1

 

It is therefore true that probability that three batteries of type one last longer than two of type two.

 

 

 

QUESTION SEVEN

 

If X BIN (100, 0.02), approximate P (4 < X < 6) in two ways

  1. With the CLT, including the half-point correction;
  2. With the Poisson approximation;
  3. With the exact formula.

 

Solution

 

(a)

 

If X BIN (100, 0.02), approximate P (4 < X < 6)

 

Therefore,

P (4 ≤ Y< 6) = P (2.5 ≤ Y≤ 5.5),

Since P (y =6) is not in the required range,

It follows that

(2.5 – 5)/ 100 ≤ (y-5)/ 100 ≤ (5.5-5)/ 100

 

= (0.426) (-1.713)

 

= 0.0281 = 0.3440

 

From the distribution table with CLT, P (4 < X < 6) is 0.3440

 

(b)

 

Using the Poisson approximation,

In general, the Poisson is given by;  W

 

If X BIN (100, 0.02), approximate P (4 < X < 6)

 

Therefore, P (4 < X < 6) = p (4.5≤ x ≤6.5)

= p({4.5 – 100}/ ) ≤ {X-100}/  ≤ {6.5- 100}/

= Ø (-9.55)- Ø (-9.35)

P = 0.3457

(C) Using the exact formula;

= p({6.5 – 100}/ ) ≤ {X-100}/  ≤ {4.5- 100}/

= Ø (-7.65)- Ø (-6.34)

P = 0.3312

QUESTION EIGHT

 

A candy maker produces mints that have a label weight of 15 grams. Assume that the distribution of the weights is N (15.5, 0.16). Assume 100 mints are sampled independently and weighted. Let Y be the number of mints that weigh less than 15 grams. Approximate < Y < 15).

 

 

Let Y be the total weight of one mint.

 

P (X < 15)

 

Therefore, P (X < 15) = P ({X – µ}/ ǫ) ≤ (15 -µ)/ ǫ

 

= P(X-15.5)/  (15-15.5)/

= Ø (-1.25)

= 1- Ø (-1.25)

= 0.1123

 

 

QUESTION NINE

P 10.

A chest has three drawers containing silver and gold coins.

  • Drawer number one has 2 gold and 3 silver coins. – Drawer number two has 2 gold and 2 silver coins.
  • Drawer number three has just two gold coins.

We select one of the drawers at random and then pick randomly one of the coins in that drawer. Given that the coin turned out to be gold, what is the probability that we chose the first drawer?

Bayes’s formula – setup is important, write down the partition, conditionals, prior probabilities

 

Solution

 

Let the drawers be represented by A, B, C. and let the chosen coins be represented by G and S for gold and silver respectively.

Using Bayes formula,

 

P(A/B) = P(A^G)/ {P(G)}

= P(A) = 1/3

 

P(G) = P(G/A) P(A) + P(B)P(G/B) + P(C)P(G/C)

= 1(1/3) +( ½) (1/3) + (0) (1/3)

= ½

Therefore, P(A/G) = (1/3) (2) = 2/3

Statistical Calculations

QUESTION TEN

P 11.

  • X has cumulative distribution function F (c) Calculate P (X > Ð.

Solution

The cumulative distribution function is given by

F(x) =

Therefore F(c) =

P (X > Ð) = p ()

= )

= p()

  • Determine the 90-th percentile of X N(—6, 4).

The 90th percentile is given by the following formular

X=μ + Zσ

X = -6 + Z4

X = -6 + 0.999

X = 5.001

  • If Y Geo (O.3). What is > 4tY > 2).

If Y Geo (0.3) therefore > 4Ty > 2

Therefore, we have > 4T (YGeo {0.3})> 2

  • 4T (YGeo {0.3})> 2

 

  • If X Poisson(l), calculate – 1) 2 ].

Poisson is given by,

W

Therefore, (- 1) 2 ]. = 1 = ∞

 

  • Xl Bin(3, 1/2) and X2 Bin(4, 1/2) are independent. Determine P(XI + = 1).

Xl Bin(3, 1/2) and X2 Bin(4, ½)

P(XI + = 1)

P(3, 1/2) and P(4, ½))

P(XI + = 1) = 0.772

  • What is the mean and variance of X — 2Y if X N (—1, 1) are independent?

Mean = E[X — 2Y] = E[X] – 2E[Y]

=E[-1]- 2E[1]

= 0-2

= -2

  • What interval describes within three standard deviations from the mean for X N (1, 0.25)?

Interval is 0.25 < X < 1

  • Find without a table.
  • We flip a fair coin 10 times. Which is more likely?

HHTHHTHHHT, TTTHHHHHHH or TTTTTTTHHH ?

The most probable outcome is HHTHHTHHHT

  • We flip a coin until a Head (H) turns up. While experimenting, we obtain the data 5, 7, 5, 5, 3, 6, 5, 6, 7, 4 equal to the number of coin flips needed to turn up the first H. Estimate the probability p of turning H for this coin.

 

Total number of outcome = 5+ 7+5+5+3+6+5+6+7+4 = 53

Therefore,

Statistical Calculations

Probability of obtaining the head = 1/2

P(H) = (1/53) (1/2) = 1/106 = 0.0094

 

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