Normal Distribution

# Normal Distribution - P Z ≤ = 1 – 0.8413 = 0.1587 Thus...

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1200 males students average height was 5 feet 7 inches with a std. Deviation of 5 inches. If heights are distributed normally what is the expected number of students that are. A. Greater than 6 feet tall b. Between 5 feet and 6 feet. Let X denotes the height of students in inches. Here it is given that X is normally distributed with mean μ = 5 feet 7 inches = 67 inches and standard deviation σ =5 inches. Then, we have 67 5 X Z - = follows a Standard Normal distribution. a) We have, P[Height of a Student is greater than 6 feet] = P[Height of a Student is greater than 72 inches] = [ 72] P X > = 67 72 67 5 5 X P - - > = [ 1] P Z > = 1 – [ 1]

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Unformatted text preview: P Z ≤ = 1 – 0.8413 = 0.1587 Thus, the expected number of students that are greater than 6 feet tall = 1200*0.1587 = 190.44 = 191 b) Also, P[Height of a Student is between 5 feet and 6 feet] = P[Height of a Student is between 60 inches and 72 inches] = P [ 60 < X < 72] = 60 67 67 72 67 5 5 5 X P---  < <     = P [ -1.4 < Z < 1] = P [ Z < 1] – P [ Z < -1.4] = 0.8413 – 0.0808 = 0.7605 Thus, the expected number of students that are between 5 feet and 6 feet = 1200*0.7605 = 912.6 = 913 Note that using MS Excel P[Z<a] can be calculated using the Excel formula =NORMSDIST(a)...
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Normal Distribution - P Z ≤ = 1 – 0.8413 = 0.1587 Thus...

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