The lengths of adult males' hands are normally distributed with mean 192 mm and standard deviation is 7.4 mm. Suppose that 40 individuals are randomly chosen. Round all answers to 4 where possible.What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)For the group of 40, find the probability that the average hand length is more than 193. Find the first quartile for the average adult male hand length for this sample size.  Round to 2 decimal places. For part b), is the assumption that the distribution is normal necessary? NoYes

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Asked Nov 18, 2019
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The lengths of adult males' hands are normally distributed with mean 192 mm and standard deviation is 7.4 mm. Suppose that 40 individuals are randomly chosen. Round all answers to 4 where possible.

  1. What is the distribution of ¯xx¯? ¯xx¯ ~ N(,)
  2. For the group of 40, find the probability that the average hand length is more than 193. 
  3. Find the first quartile for the average adult male hand length for this sample size.  Round to 2 decimal places. 
  4. For part b), is the assumption that the distribution is normal necessary? NoYes
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Expert Answer

Step 1

Central Limit Theorem for mean:

If a random sample of size n is taken from a population having mean µ and standard deviation σ then, as the sample size increases, the sample mean approaches the normal distribution with mean µ and standard deviation σ/sqrt(n).

Step 2

1.

Finding the distribution of X-bar:

It is given that the lengths of adults’ male hands follows normal distribution with mean 192mm and standard deviation 7.4mm.

That is, µ= 192, σ = 7.4.

Random samples of 40 (n) individuals is considered.

Denote the random variable X as the length of hand for a randomly selected individual.

By central limit theorem for mean, the average hand length follows a normal distribution with mean μ = 192 and standard deviation 7.4/sqrt (40) = 1.1700.

Thus, the distribution of X-bar ~ N(192,1.17).

Step 3

2.

Finding the probability that the average hand length is more than 193:

The probability that the aver...

P(X>193)1-P(X< 193)
=1-0.8036
Using the EXCEL formulae:
NORM.DIST (193,192,1.17,1)
-0.1964
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P(X>193)1-P(X< 193) =1-0.8036 Using the EXCEL formulae: NORM.DIST (193,192,1.17,1) -0.1964

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