Class 04. Wunderdog and the Normal Distribution. EMBS Section 6.2. Class 03 Assignment. Answers are posted on the course website My office hours are IN THE CLASSROOM. 3 to 430 on class days Or email me for an appointment [email protected] TA Office hours Sundays and Tuesday Nights
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Class 04. Wunderdog and the Normal Distribution
EMBS Section 6.2
Binomial pmf with n=149, p=0.5
Each possible outcome x has a mass of probability calculated as BINOM.DIST(x,149,.5,false)
All three are “bell-shaped curves”
P=0.5
P=0.8
P=0.2
EMBS Fig 6.4, p 249
Just like the binomial, the normal is a FAMILY of distributions. The member of the Normal family we want to use is the one with the mean and standard deviation that match our binomial.
Mean=E(X)=74.5
Standard deviation = 6.1
Normal with μ=74.5, σ=6.1
P(x≥87) =
1-BINOMDIST(86,149,.5,true)
=0.024
X is discrete
P(x=87) = 0.008
P(x≥87) =
P(x>87) =
1-NORMDIST(87,74.5,6.1,true)
=0.020
X is continuous
P(x=87) = 0
Weights of CEO’s are normally distributed with µ = 155 and σ=25. What percentage of CEO’s do we expect weigh between 160 and 200?
=NORMDIST(200,155,25,true)-NORMDIST(160,155,25,true)
= 0.964 – 0.579
= 0.385
EMBS problem 21, page 260
A person must score in the top 2% of the population on an IQ test to qualify for membership in MENSA (U.S. Airways Attache, September 2000). If the population of IQ scores is normal with mean of 100 and standard deviation of 15, what score qualifies one for MENSA?
We want the score, x, such that the probability(X<x) is 0.98.
=NORMINV(.98,100,15)
= 130.8
EMBS Fig 6.4, p 249
z tells us where x is on its normal curve.
z is how far x is above/below the mean in units of standard deviation.
z is all we need to answer a probability quesgtion.
The standard normal distribution.
Uses z as the input. We needed calculate the z in order to answer probability questions.
=NORMDIST(x1,μ,σ,true)