The following data give the ages (in years) of all six members of a family. 57 5

Question

The following data give the ages (in years) of all six members of a family.

57

53

31

27

22

14

Let x denote the age of a member of this family. What is the probability P(x=22)?

Round your answer to three decimal places.

P(x=22)=

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Consider a large population with =70 and =12. Assuming nN0.05, find the mean and standard deviation of a sample mean, x¯, for a sample size of 20.

Round your answers to three decimal places.

x¯=

57

53

31

27

22

14

Explanation / Answer

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/ }] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or using Excel Function……………………………………..(5)

Now, to work out solution,

X = age of a member of the family. We assume age follows Normal Distribution with mean µ and variance 2. Since these values are not given, we will estimate these with sample mean and sample variance. From the given data, µ = 34 and variance 2 = 248.6667 or 2 = 15.769.

Q1

For a continuous variable like age, probability at a particular value of X is not defined since the probability function, f(x) gives the probability of not x, but the nprobability of x to x + dx. Also, when we say the age is 22, it could be anything from 21.5 to 22.5, which on rounding becomes 22. So, P(x = 22) = P(21.5 X 22.5)

= P[{21.5 - 34}/15.769} Z {22.5 - 34}/15.769}] [vide (2) under Back-up Theory]

= P(- 0.793 Z 0.729)

= P(Z 0.729) - P(Z - 0.793) = 0.7670 – 0.2139 [using Excel Function]

= 0.5531 ~ 0.553 (Rounded to three decimal places) ANSWER

Q2

By the given condition, X can be assumed to have Normal Distribution.

[vide (2) under Back-up Theory], mean of Xbar = µ = 70, of Xbar = /n = 12/20

= 2.683

So, µXbar = 70 and Xbar = 2.683 ANSWER

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