Question

You have information to suggest that a certain continuous
variable of a population has a mean of μ=12.44μ=12.44 and a
standard deviation of σ=6.78σ=6.78. You are to randomly pick
n=68n=68 individuals from this population and observe the value of
the population variable on **each** . This value is to
measured as XiXi.

After the random sample of n=68n=68 has been taken, you are asked
to consider the behavior of the statistic X¯¯¯¯X¯.

**(a)** Complete the statement below. Use at least two
decimals in each numeric answer.

The distribution of X¯¯¯¯X¯ ? is approximately Normal is
exactly Normal is skewed to the right has an unknown
shape with a mean μX¯¯¯¯¯=μX¯=

and a standard deviation σX¯¯¯¯¯=σX¯=

**(b)** Find the probability that mean of the sample
of n=68n=68 is between 11.94 and 12.84. Use at least three decimals
in your zz-values, and enter your answer to four decimal

Answer #1

(a) Since we do not know if the population from which the sample is selected, but also that n > 30, therefore

The distribution of X is **approximately normal** with a
mean
= **12.44** and a
standard deviation
= 6.78 / sqrt(68) = **0.82**

**___________________________________________________________**

(b) To find P (11.94 < X < 12.84) = P(X < 12.84) – P(X < 11.94)

For P( X < 12.84)

Z = (12.84 – 12.44) / 0.82 = 0.488

The probability for P(X < a) from the normal distribution tables is = 0.6872

For P( X < 11.94)

Z = (11.94 – 12.44) / 0.82 = -0.610

The probability for P(X < 11.94) from the normal distribution tables is = 0.2709

**Therefore the required probability is 0.6872 – 0.2709 =
0.4163**

**____________________________________________________________**

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