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View History Bookmarks Tools Window Help MAT181-04 Fall Leland Statisties y chiries 1027/17 3.51 PM Test: Chapter 3 Post-Test Submit Teet This Question: 1 pt 19 of 27 (11 completel This Test: 27 pts possible oh Heights of men on a basebal team have a approximate percentage of the men between the following values? a. 163 and 187cm b. 169 om and 181 cm bell shaped distribution with a mean of 175 am and a standard deviation of 6om Using the empirical nule, what is the %ofthe men are between 163 cm and 187cm. Round to one decimal place as needed.) b.% Round to one decimal place as needed.) of tho men are between 169cm and 181 cm Enter your answer in each of the answer boxes

Explanation / Answer

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 ) = 175
standard Deviation ( sd )= 6
a.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 163) = (163-175)/6
= -12/6 = -2
= P ( Z <-2) From Standard Normal Table
= 0.0228
P(X < 187) = (187-175)/6
= 12/6 = 2
= P ( Z <2) From Standard Normal Table
= 0.9772
P(163 < X < 187) = 0.9772-0.0228 = 0.9545
95.5% is b/w 163 < X < 187
b.
To find P(a < = Z < = b) = F(b) - F(a)
P(X < 169) = (169-175)/6
= -6/6 = -1
= P ( Z <-1) From Standard Normal Table
= 0.1587
P(X < 181) = (181-175)/6
= 6/6 = 1
= P ( Z <1) From Standard Normal Table
= 0.8413
P(169 < X < 181) = 0.8413-0.1587 = 0.6827
68.3% is b/w 169 < X < 181

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