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MID-TERM EXAM # 2(Practice)

Prof. Deoki N. Sharma

401 Basic Statistics for Research

Name: ____________________

Net ID# Date: ________________

NOTE: PLEASE READ CAREFULLY, SHOW KEY STEPS IN YOUR WORK.

1. A surgical technique is performed on seven patients. You are told there is a

70% chance of success. Find the probability that the surgery is successful for

(a) exactly five patients, (b) at least five patients, and (c ) less than five

patients.

(d) Determine the mean, variance and standard deviation of the random

variable X.

Let x is number of patients.

Solution

This is a Binomial Problem with n = 7 and P(success) = 0.7

a ) exactly 5 patients

P(x = 5) = 7C5*0.7^5*0.3^2 = 0.3177

b) at least 5 patients

P(5<= x <=7) = P(x = 5) + P(x = 6) + P(x = 7 )

= 7C5*0.7^5*0.3^2 + 7C6*0.7^6*0.3^1 + 7C7*0.7^7*0.3^0

= 0.3177 + 0.2471 + 0.0824

= 0.6471

c) less than 5 patients

P(0<= x < 5) = P(x = 0) + P(x = 1) + P(x = 2) + P(x = 3) + P(x = 4)

= 7C0*0.7^0*0.3^7 + 7C1*0.7^1*0.3^6 + 7C2*0.7^2*0.3^5 +

7C3*0.7^3*0.3^4 + 7C4*0.7^4*0.3^3

= 0.0002 + 0.0036 + 0.0250 + 0.0972 + 0.2269

= 0.3529

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d) The mean of a binomial distribution is µ=np

Therefore, µ= 7*0.7 = 4.9

Variance of a binomial distribution is np(1-p)

Therefore, variance = 7*0.7*(1-0.7) = 4.9*0.3 = 1.47

Standard deviation = √1.47 = 1.21

2. The mean number of bankruptcies filed per minute in the United States in a

recent year was about two. Find the probability that:

(a) exactly five businesses will file bankruptcy in any given minute. Let x is

number of business.

This is a Poisson distribution of data with λ= 3 minutes

P(5) = (λ^5 * e^- λ) /5!

= (3^5 * e^-3) /5!

= 0.1008

(b) At least five businesses will file bankruptcy in any given minute.

P(X>=5) = 1 – [ P(0) + P(1) + P(2) + P(3) + P(4) ]

= 1- [ (λ^0 * e^- λ) /0! + (λ^1* e^-λ) /1! + (λ^2 * e^- λ)/2! +

(λ^3* e^-λ) /3! + (λ^4 * e^- λ)/4! ]

= 1 – [(3^0 * e^-3) /0! + (3^1 * e^-3) /1! + (3^2 * e^-3) /2! +

(3^3 * e^-3) /3! + (3^4 * e^-3) /4! ]

= 1- [ 0.0498 + 0.1493 + 0.2240 + 0.2240 + 0.1680 ]

= 0.1847

( c) More than five businesses will file bankruptcy in any given minute

P(X> 5) = P(X>=5) – P(5)

= 0.1847 – 0.1008

= 0.0839

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3. Given the following probability distributions:

(a) Compute the expected value for the distribution: E(x) = )(xpx

E(x) = )(xpx = 2.10

(b) Compute the standard deviation for the distribution. Hint:

222 )( xpx

222 )( xpx = 6 – (2.10)^2 = 1.59

59.1 = 1.26

X

P(x)

X.P(x)

)(2 xPX

0 .1

1 .25

2 .30

3 .15

4 .20

)(. xPX

)(.2 xPX

X

P(x)

X.P(x)

)(2 xPX

0 .1 0 0

1 .25 0.25 0.25

2 .30 0.60 1.20

3 .15 0.45 1.35

4 .20 0.80 3.20

)(. xPX 2.10

)(.2 xPX 6.0

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4. Scores on a certain nationwide college entrance examination follow a

normal distribution with a mean of 500 and a standard deviation of 100. Find

the probability that a student will score: Given a normal distribution

with ,100500 and {Hint: Z =

x }

(a). What is the probability that a student will score Over 650 i.e. P(X > 650) =

Z = (650 – 500)/ 100 = 1.5

Using z-score index tables;

P(over 650) = 1 - 0.9332 = 0.0668

(b). What is the probability that a student will score less than 250 i.e. P(x<250)

Z = (250 – 500)/ 100 = -2.5

Using z-score index tables;

P(less than 250) = 0.0062

(c ). What is the probability that a student will score between 325 and 675 i.e.

P(325 < X < 675)

Z @ 325 = (325 – 500)/ 100 = -1.75

P(less than 325) = 0.040059

Z @ 675 = (675-500)/100 = 1.75

P(less than 675) = 0.959941

P(between 325 and 675) = 0.959941 - 0.040059 = 0.919882 = 0.9199

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5. On a dry surface, the braking distance ( in meters) of a certain car is a

normal distribution with mean 45.1 m and standard deviation 0.5.

(a) Find the braking distance that corresponds to z = 1.8

Z =1.8 = (x – 45.1)/ 0.5

0.9 = x – 45.1

X = 45.1 + 0.9 = 46

Required braking distance = 46 m

(b) Find the braking distance that represents the 91st percentile.

The Z score associated with 91st percentile is Z = 1.3408

Using the same calculations as above

1.3408 = (x – 45.1) / 0.5

0.6704 = x – 45.1

X = 45.77

Required braking distance = 45.77 m

( c) Find the z-score for a braking distance of 46.1 m.

Z = (46.1 – 45.1 ) / 0.5 = 1/0.5 = 2

(d) Find the probability that the braking distance is less than or equal to 45 m.

Z = (45 – 45.1)/0.5 = -0.2

Using z-score index tables;

P(X<=45) = 0.4207

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(e ) Find the probability that the braking distance is greater than 46.8 m.

Z = (46.8 – 45.1)/0.5 = 3.4

Using z-score index tables;

P(X > 46.8) = 1 – 0.9997 = 0.0003

( f ) Find the probability that the braking distance is between 45 m and 46.8 m.

P (45 < X <46.8) = 0.9997 – 0.4207 = 0.579

6. A business wants to estimate the true mean annual income of its customers.

It randomly samples 220 of its customers. The mean annual income was

$61,400 with a standard deviation of $2,200. Find a 95% confidence interval

for the true mean annual income of the business’ customers.

Solution

n=220

xbar=61400

s=2200

alpha=0.05

First of all, we know the distribution is normal, since the sample size is large

(>30). That means our test statistic is either Z or t. Since the standard

deviation was estimated from a sample, the population standard deviation is

unknown. This means we will use the t-statistic.

So, the confidence interval will be of the form:

mean +/- t(1-alpha/2;n-1) x s/sqrt(n)

ie. 61400 +/- t(0.975;219) x 2200/sqrt(220)

ie. 61400 +/- 0.8347 x 148.324

ie. 61400 +/- 124

ie. (61276,61524)

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7. A business wants to estimate the true mean annual income of its customers.

The business needs to be within $250 of the true mean. The business

estimates the true population standard deviation is around $2,400. If the

confidence level is 90%, find the required sample size in order to meet the

desired accuracy.

Solution

m = margin of error = 250

standard deviation = 2400

90% confidence level

we need z*, at 90%, z* is 1.645

n = (1.645*2400/250)^2 = 249.39 and we will always round up to 250.

At least 250 samples.

8. The weight of an almond is normally distributed with mean 0.05 ounce and

standard deviation 0.015 ounce. Find the probability that a package of 100

almonds weighs between 4.8 and 5.3 ounces. That is, find the probability

that X is between 0.048 and 0.053 ounce. Hint: Z =

n

X

}

Each almond is a weight from a normal distribution. Therefore the 100

almonds come from 100 such normal distributions. Now when you add

distributions you add the means and variances (note variances NOT standard

deviations).

Therefore the packet mean will be 100*0.05 = 5 ounces.

The individual variance is 0.015^2 so the packet variance is 100*(0.015^2) =

0.0225 so stan. dev. = sqrt(0.0225) = 0.15

Now, Converting to Z Scores

Z @ 4.8 = (4.8 – 5 ) /0.15 = -1.33

Z @ 5.3 = (5.3 – 5 )/ 0.15 = 2

Required probability = 0.9772 – 0.0912 = 0.886