Question

Suppose that a random sample of size 100 is to be selected from a population with mean 50 and standard deviation 8. What is the probability that a sample mean will be greater than 50.8? Round your answer to three decimal places.

Answer 1

Answer:

0.159

Step-by-step explanation:

Given that:

Sample size (n) = 100

Population mean(pm) = 50

Standard deviation (s) = 8

Probability that Sample mean (m) will be greater Than 50.8

Using the relation :

(sample mean - population mean) / (standard deviation /sqrt(n))

P(m > 50.8)

Z = (50.8 - 50) / (8/ sqrt(100))

Z = 0.8 / (8/ 10)

Z = 0.8 / 0.8

Z = 1

P(Z > 1) = 0.15866 ( Z probability calculator)

Hence,

P(Z > 1) = 0.159 ( 3 decimal places)

Answer 2

The probability that a sample mean will be greater than 50.8 is P(Z > 1) = 0.159.

Given that,

Suppose that a random sample of size 100 is to be selected from a population with a mean of 50 and a standard deviation of 8.

We have to determine,

What is the probability that a sample mean will be greater than 50.8?

According to the question,

Sample size (n) = 100

Population mean(pm) = 50

Standard deviation (s) = 8

The probability that Sample mean (m) will be greater than 50.8.

Therefore,

$\dfrac{(sample \ mean - population \ mean) \times \sqrt{n}}{standard \ deviation}$

Where P(m > 50.8)

$Z = \dfrac{50.8 - 50 \times \sqrt{100}}{8}\\\\Z =\dfrac{ 0.8 \times 10 }{8}\\\\Z = \dfrac{0.8 }{0.8}\\\\Z = 1$

P(Z > 1) = 0.15866 ( Z probability calculator)

Hence,  The probability that a sample mean will be greater than 50.8 is P(Z > 1) = 0.159.

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