Chapter 7 instructions I help Question 2 (of 5 E| | save & Exit | Submit 2.00 po

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Chapter 7 instructions I help Question 2 (of 5 E| | save & Exit | Submit 2.00 points The mean of a normal probability distribution is 320; the standard deviation is 8. a. About 68% of the observations lie between what two values? Value 1 Value 2 b. About 95% of the observations lie between what two values? Value 1 Value 2 c. Practically all of the observations lie between what two values? Value 1 Value 2 Worksheet Difficulty: 1 Basic Learning Objective: 07-03 Describe the standard normal probability distribution and use it to calculate probabilities.

Explanation / Answer

2.

NORMAL DISTRIBUTION
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 ) = 320
standard Deviation ( sd )= 8
a.
About 68% of the area under the normal curve is within one standard deviation of the mean. i.e. (u ± 1s.d)
So to the given normal distribution about 68% of the observations lie in between
= [320 ± 8]
= [ 320 - 8 , 320 + 8]
= [ 312 , 328 ]
b.
About 95% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 2s.d)
So to the given normal distribution about 95% of the observations lie in between
= [320 ± 2 * 8]
= [ 320 - 2 * 8 , 320 + 2* 8]
= [ 304 , 336 ]
c.
practically all values between two is
About 99.7% of the area under the normal curve is within two standard deviation of the mean. i.e. (u ± 3s.d)
So to the given normal distribution about 99.7% of the observations lie in between
= [320 ± 3 * 8]
= [ 320 - 3 * 8 , 320 + 3* 8]
= [ 296 , 344 ]

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