Question

Suppose an x distribution has mean μ = 4. Consider two corresponding

x

distributions, the first based on samples of size n = 49 and the second based on samples of size n = 81.

(a) What is the value of the mean of each of the two

x

distributions?

For n = 49, μ x =

For n = 81, μ x =

Suppose the heights of 18-year-old men are approximately normally distributed, with mean 70 inches and standard deviation 5 inches.

(a) What is the probability that an 18-year-old man selected at random is between 69 and 71 inches tall? (Round your answer to four decimal places.)


(b) If a random sample of twenty-nine 18-year-old men is selected, what is the probability that the mean height x is between 69 and 71 inches? (Round your answer to four decimal places.)

Let x be a random variable that represents white blood cell count per cubic milliliter of whole blood. Assume that x has a distribution that is approximately normal, with mean μ = 6500 and estimated standard deviation σ = 2100. A test result of x < 3500 is an indication of leukopenia. This indicates bone marrow depression that may be the result of a viral infection.

(a) What is the probability that, on a single test, x is less than 3500? (Round your answer to four decimal places.)


(b) Suppose a doctor uses the average x for two tests taken about a week apart. What can we say about the probability distribution of x?

The probability distribution of x is not normal.The probability distribution of x is approximately normal with μ_x = 6500 and σ_x = 1050.00.    The probability distribution of x is approximately normal with μ_x = 6500 and σ_x = 1484.92.The probability distribution of x is approximately normal with μ_x = 6500 and σ_x = 2100.

What is the probability of x < 3500? (Round your answer to four decimal places.)


(c) Repeat part (b) for n = 3 tests taken a week apart. (Round your answer to four decimal places.)

Answer 1

Ans:

1)Sampling distribution of sample means will be normal distribution with mean equals to population mean.

For n=49

For n=81

2)

a)

z(69)=(69-70)/5

z(71)=(71-70)/5

P(-0.2<z<0.2)=P(z<0.2)-P(z<-0.2)

=0.5793-0.4207=0.1586

b)

z(69)=(69-70)/(5/sqrt(29))=1.077

z(71)=(71-70)/(5/sqrt(29))=1.077

P(-1.077<z<1.077)=P(z<1.077)-P(z<-1.077)

=0.8592-0.1407=0.7185