6. Men\'s heights are normally distributed with a mean of 69.0 inches and a stan

Question

6. Men's heights are normally distributed with a mean of 69.0 inches and a standard deviation of 2.8 inches. The height of the doorway to a cave is 68.0 inches. a) What percentage of men can fit through the doorway without bending? 4ps Round your answer to the nearest tenth percent b) If a random sample of 49 men is selected, find the probability that the mean will be less than 68.0 inches. Write your answer as a decimal rounded to 4 decimal places 4 pts) 7. Suppose you want to find the 90% confidence interval for a population mean using data from a sample of size 35, where the population standard deviation is not known. Find the critical value t Round your answer to 3 decimal places. 2 pts)

Explanation / Answer

Q6.

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 69
standard Deviation ( sd )= 2.8
a.
P(X < 68) = (68-69)/2.8
= -1/2.8= -0.3571
= P ( Z <-0.3571) From Standard Normal Table
= 0.3605
b.
mean of the sampling distribution ( x ) = 69
standard Deviation ( sd )= 2.8/ Sqrt ( 49 ) =0.4
sample size (n) = 49
P(X < 68) = (68-69)/2.8/ Sqrt ( 49 )
= -1/0.4= -2.5
= P ( Z <-2.5) From Standard NOrmal Table
= 0.00621

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