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 μx = 6500 and σx = 1050.00. The probability distribution of x is approximately normal with μx = 6500 and σx = 1484.92.The probability distribution of x is approximately normal with μx = 6500 and σx = 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.)
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
Suppose an x distribution has mean μ = 4. Consider two corresponding x distributions, the first...
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