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.

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,

Where P(m > 50.8)

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.

To know more about Probability click the link given below.

link

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)

total amount of money that he should pay if he has not used his duty free exemption is $98

step-by-step explanation:

given a united states citizen returning to the states declares the following items at the customs office: 3 shirts at $8.50 each 2 dresses at $27.50 each 1 pair of gold cuff links at $17.50 per pair. we have to find how much he pay

cost of 3 shirts if one cost at $8.50 is

cost of 2 dresses if one cost at $27.50 is  

cost of 1 pair of gold cuff at $17.50

so, total amount of money that he should pay if he has not used his duty free exemption is   $25.5+$55+$17.50=$98