Cuotion 17522 marks n For the given frequency distribution

Cuotion / 17,5,2.2 marks) n For the given frequency distribution ass limits 24 -29 Fregnency 12 20 a) Calculate the sample standard devtation s of the ahove b) Calculate the median, of the above distribution? 2) If A and B are two events in S. I P(A)-0.5, P(B) 07 and P(AuB) 0.65, find the probabilities of a) P(AnB) b) P(B1A)

We are given that frequency distribution.

From frequency distribution we have to find standard deviation and median.

Sample standard deviation:

s^2 = 1/n-1* ( f*X² - n*mean² )

where n = f

f is frequency.

X is mid point.

mean = f*x / f

Find n/2^th observation and take cumulative frequency.

Cumulative Frequency corresponding to a particular value is the sum of all the frequencies up to and including that value.

Median = L + [((n/2) - B) / G]*w

where, L is the lower class boundary of the group containing the median

The complete table in EXCEL is,

mean = 677 / 50 = 13.54

s^2 = 1/50-1 * (1128.5 - 50*13.54²) = 43.1412

s = sqrt(43.1412) = 6.5682

sample standard deviation = 6.5682

n/2 = 50/2 = 25

25 is lie in the 12-17 class.

So median class is 12-17.

Median = 11.5 + [ ((50/2) - 18) / 20 ] * 6 = 13.6

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A and B are two events.

P(A) = 0.5

P(Bbar) = 0.7

P(B) = 1 - P(Bbar) = 1 - 0.7 = 0.3

P(A U B ) = 0.65

By using addition law,

P(A U B) = P(A) + P(B) - P(A B)

0.65 = 0.5 + 0.3 - P(A B)

P(A B) = 0.8 - 0.65 = 0.15

P(B / A) = P(A B) / P(A) = 0.15 / 0.5 = 0.3

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90% confidence interval for mean

This question we can done using TI-83 calculator :

Here we use t-interval because population standard deviation is unknown and sample data is given.

steps :

STAT --> ENTER --> enter all the data values --> STAT --> TESTS --> 8:TInterval --> ENTER --> highlight on data --> ENTER --> List :L1 --> Freq :1 --> C-level = 0.90 --> Calculate --> ENTER

90% confidence interval for mean is (7.8592,11.141)

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Here we have to test the hypothesis that,

H0 : mu = 30 Vs H1 : mu > 30

alpha = 0.005 = level of significance.

Xbar = 30.8

S = 1.8

n = 32

mu0 = 30

Here also we use t-test because population standard deviation is not given.

By using TI-83 calculator steps are,

STAT --> TESTS --> 2: T-Test --> ENTER --> highlight on Stats --> ENTER --> Enter mu0, Xbar, Sx (that is S) and n --> select > mu0 --> ENTER --> Calculate --> ENTER

Output is,

t = 10.969655

P-value = 1.1583E-5 = 0.00001

P-value < alpha

Reject H0 at 0.005 level of significance.

Conclusion : Population mean is greator than 30.

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X is normally distributed.

= 2

P(X > 6) = 0.9772

Convert 6 into z-score.

z = (6 - mu) / = (6 - mu) / 2

That is P(Z > (6-mu) / 2) = 0.9772

By using EXCEL we have to use inverse normal probability by using syntax,

=NORMSINV(probability)

where probability = 0.9772

This will gives answer 1.9991.

Compare it with (6-mu)/2.

That is (6 - mu) / 2 = 1.9991

6 - mu = 3.9982

mu = 6 - 3.9982 = 2.0018

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Given that X is a binomial random variable with n=50 and p = 0.2

We have to find P(X > 14) using normal approximation.

mean = n*p = 50*0.2 = 10

variance = np(1-p) = 50*0.2*(1-0.2) = 8

sd = sqrt(variance) = sqrt(8) = 2.8284

P(X > 14):

Convert 14 into z-score.

z = (X - mu) / sd = (14 - 10) / 2.8284 = 1.4142

Now we have to find P(Z > 1.4142).

EXCEL syntax :

1 - NORMSDIST(z) (because probability is right sided)

where z is test statistic value.

P(Z > 1.4142) = 0.0786

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X is a binomial random variable with mean =2 ,variance = 1.6.

And we have to find P(X = 4).

We know that, mean = np = 2

variance = npq = 1.6

divide varinance by mean,

npq / np = 1.6 / 2

q = 0.8

p = 1 - q = 1 - 0.8 = 0.2

np = 2

n*0.2 = 2

n = 2 / 0.2 = 10

Therefore, n = 10, p=0.2, q = 0.8