LECTURE 6 DISTRIBUTIONS

LECTURE 6 DISTRIBUTIONS

OVERVIEW Uniform Distribution Normal Distribution Random Variables Continuous Distributions MOST OF THE SLIDES ADOPTED FROM OPENINTRO STATS BOOK.

NORMAL DISTRIBUTION Unimodal and symmetric, bell shaped curve Many variables are nearly normal, but none are exactly normal Denoted as standard deviation -> Normal with mean and

HEIGHTS OF MALES Male of heights almost normally distributed Shift: People adds a couple of inches to their heights

NORMAL DISTRIBUTION Mean = 0-1 0 1

NORMAL DISTRIBUTION Mean = 0-1 0 1

NORMAL DISTRIBUTION Mean = 0-1 0 1

NORMAL DISTRIBUTION Mean = 0 standard deviation -1 0 1

NORMAL DISTRIBUTION Mean = 0 standard deviation -1 0 1

NORMAL DISTRIBUTION Mean = 0 standard deviation -1 0 1

NORMAL DISTRIBUTION Mean = standard deviation -1 0 1

NORMAL DISTRIBUTION Mean = 1 standard deviation -1 0 1

NORMAL DISTRIBUTION Mean = 1 standard deviation -1 0 1

EXAMPLE SAT scores are distributed nearly normally with mean 1500 and standard deviation 300. ACT scores are distributed nearly normally with mean 21 and standard deviation 5. A college admissions officer wants to determine which of the two applicants scored better on their standardised test with respect to the other test takers: Pam, who earned an 1800 on her SAT, or Jim, who scored a 24 on his ACT?

EXAMPLE (CONTINUED) Since we cannot just compare these two raw scores, we instead compare how many standard deviations beyond the mean each observation Pam s score: is. Jim s score: = standard dev. above the mean. standard dev. above the mean.

STANDARD Z-SCORE Z score of an observation is the number of standard deviations it falls above or below the mean Z scores are defined for distributions of any shape, but only when the distribution is normal can we use Z scores to calculate percentiles Observations that are more than 2 SD away from the mean ( Z >2) are usually considered unusual.

PERCENTILES Percentile is the percentage of observations that fall below a given data point. Graphically, percentile is the area below the probability distribution curve to the left of that observation.

CALCULATING PERCENTILES

EXAMPLE At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of ketchup in the bottle is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles have less than 35.8 ounces of ketchup?

EXAMPLE At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of ketchup in the bottle is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles have less than 35.8 ounces of ketchup?

EXAMPLE At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of ketchup in the bottle is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles have less than 35.8 ounces of ketchup? Let x= amount of ketchup in a bottle:

EXAMPLE At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of ketchup in the bottle is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles have less than 35.8 ounces of ketchup? Let x= amount of ketchup in a bottle:

EXAMPLE At Heinz ketchup factory the amounts which go into bottles of ketchup are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. Once every 30 minutes a bottle is selected from the production line, and its contents are noted precisely. If the amount of ketchup in the bottle is below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles have less than 35.8 ounces of ketchup? Let x= amount of ketchup in a bottle:

EXAMPLE Z=-1.82

EXAMPLE Z=-1.82

EXAMPLE Z=-1.82

EXAMPLE Z=-1.82

EXAMPLE..are supposed to be normally distributed with mean 36 oz. and standard deviation 0.11 oz. must not be below 35.8 oz. or above 36.2 oz., then the bottle fails the quality control inspection. What percent of bottles pass?

CUTTOFF POINTS Body temperatures of healthy humans are distributed nearly normally with mean 98.2F and standard deviation 0.73F. What is the cutoff for the lowest 3% of human body temperatures?

CUTTOFF POINTS Body temperatures of healthy humans are distributed nearly normally with mean 98.2F and standard deviation 0.73F. What is the cutoff for the lowest 3% of human body temperatures?

CUTTOFF POINTS Body temperatures of healthy humans are distributed nearly normally with mean 98.2F and standard deviation 0.73F. What is the cutoff for the lowest 3% of human body temperatures?

CUTTOFF POINTS Body temperatures of healthy humans are distributed nearly normally with mean 98.2F and standard deviation 0.73F. What is the cutoff for the lowest 3% of human body temperatures?

68-95-99 RULE about 68% falls within 1 SD of the mean, about 95% falls within 2 SD of the mean, about 99.7% falls within 3 SD of the mean

EXERCISE

EXERCISE Assume that in a city, the heights of people are distributed normally with a mean of 170 meters and a standard deviation of 5 cm. What is the expected number of people having a height between 165-175 meters (in terms of population percentage)?

EXERCISE Assume that in a city, the heights of people are distributed normally with a mean of 170 meters and a standard deviation of 5 cm. What is the expected number of people having a height between 165-175 meters (in terms of population percentage)? 68%

PRACTICE Which of the following is false? (a) Majority of Z scores in a right skewed distribution are negative. (b) In skewed distributions the Z score of the mean might be different than 0. (c) For a normal distribution, IQR is less than 2 x SD. (d) Z scores are helpful for determining how unusual a data point is compared to the rest of the data in the distribution.

PRACTICE Which of the following is false? (a) Majority of Z scores in a right skewed distribution are negative. (b) In skewed distributions the Z score of the mean might be different than 0. (c) For a normal distribution, IQR is less than 2 x SD. (d) Z scores are helpful for determining how unusual a data point is compared to the rest of the data in the distribution.

BINOMIAL DISTRIBUTION If p represents probability of success, (1-p) represents probability of failure, n represents number of independent trials, and k represents number of successes

PRACTICE Which of the following is not a condition that needs to be met for the binomial distribution to be applicable? (a) the trials must be independent (b) the number of trials, n, must be fixed (c) each trial outcome must be classified as a success or a failure (d) the number of desired successes, k, must be greater than the number of trials (e) the probability of success, p, must be the same for each trial

PRACTICE Which of the following is not a condition that needs to be met for the binomial distribution to be applicable? (a) the trials must be independent (b) the number of trials, n, must be fixed (c) each trial outcome must be classified as a success or a failure (d) the number of desired successes, k, must be greater than the number of trials (e) the probability of success, p, must be the same for each trial

PRACTICE A 2012 Gallup survey suggests that 26.2% of Americans are obese. Among a random sample of 10 Americans, what is the probability that exactly 8 are obese?

PRACTICE A 2012 Gallup survey suggests that 26.2% of Americans are obese. Among a random sample of 10 Americans, what is the probability that exactly 8 are obese?

PRACTICE

PRACTICE What is the probability that 2 randomly chosen people share a birthday?

PRACTICE What is the probability that 2 randomly chosen people share a birthday? Pretty low, 1/365

PRACTICE What is the probability that 2 randomly chosen people share a birthday? Pretty low, 1/365 What is the probability that at least 2 people out of 366 people share a birthday?

PRACTICE What is the probability that 2 randomly chosen people share a birthday? Pretty low, 1/365 What is the probability that at least 2 people out of 366 people share a birthday? Exactly 1! (Excluding the possibility of a leap year birthday.)

PRACTICE A 2012 Gallup survey suggests that 26.2% of Americans are obese. Among a random sample of 100 Americans, how many would you expect to be obese?

BINOMIAL DISTRIBUTION Mean and standard deviation Obesity example, We would expect 26.2 out of 100 randomly sampled Americans to be obese, with a standard deviation of 4.4.

UNUSUAL OBSERVATIONS An August 2012 Gallup poll suggests that 13% of Americans think home schooling provides an excellent education for children. Would a random sample of 1,000 Americans where only 100 share this opinion be considered unusual? a) No b) Yes

UNUSUAL OBSERVATIONS An August 2012 Gallup poll suggests that 13% of Americans think home schooling provides an excellent education for children. Would a random sample of 1,000 Americans where only 100 share this opinion be considered unusual? a) No b) Yes

UNUSUAL OBSERVATIONS An August 2012 Gallup poll suggests that 13% of Americans think home schooling provides an excellent education for children. Would a random sample of 1,000 Americans where only 100 share this opinion be considered unusual? a) No b) Yes

UNUSUAL OBSERVATIONS An August 2012 Gallup poll suggests that 13% of Americans think home schooling provides an excellent education for children. Would a random sample of 1,000 Americans where only 100 share this opinion be considered unusual? a) No b) Yes

UNUSUAL OBSERVATIONS An August 2012 Gallup poll suggests that 13% of Americans think home schooling provides an excellent education for children. Would a random sample of 1,000 Americans where only 100 share this opinion be considered unusual? a) No b) Yes

UNUSUAL OBSERVATIONS For this activity you will use a web applet. Go to http://goo.gl/viyj4w and choose Binomial coin experiment in the drop down menu on the left. Set the number of trials to 20 and the probability of success to 0.15. Describe the shape of the distribution of number of successes. Keeping p constant at 0.15, determine the minimum sample size required to obtain a unimodal and symmetric distribution of number of successes. Please submit only one response per team.

UNUSUAL OBSERVATIONS Hollow histograms of samples from the binomial model where p=0.1, and n=10,30, 100 and 300. What happens as n increases?

HOW LARGE IS LARGE ENOUGH? The sample size is considered large enough if the expected number of successes and failures are both at least 10.

NORMAL APPROXIMATION Below are four pairs of Binomial distribution parameters. Which distribution can be approximated by the normal distribution? a) n = 100; p = 0:95 b) n = 25; p = 0:45 c) n = 150; p = 0:05 d) n = 500; p = 0:015

NORMAL APPROXIMATION Below are four pairs of Binomial distribution parameters. Which distribution can be approximated by the normal distribution? a) n = 100; p = 0:95 b) n = 25; p = 0:45 c) n = 150; p = 0:05 d) n = 500; p = 0:015

NORMAL APPROXIMATION Below are four pairs of Binomial distribution parameters. Which distribution can be approximated by the normal distribution? a) n = 100; p = 0:95 b) n = 25; p = 0:45 c) n = 150; p = 0:05 d) n = 500; p = 0:015

MICHELOB VS. SCHLITZ VS. REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability VS. REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability VS. probability of least say 40% of Michelob drinkers prefer Schlitz? REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability VS. probability of least say 40% of Michelob drinkers prefer Schlitz? 98% REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ Answer = P(40 Schlitz) + P(41 Schlitz)+.. + P(100 Schlitz) P (k Schlitz) = C(100,k)0.5 k 0.5 100 k REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability, N= 100 people Answer = P(40 Schlitz) + P(41 Schlitz)+.. + P(100 Schlitz) P (k Schlitz) = C(100,k)0.5 k 0.5 100 k REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ Average person prefer with 0.5 probability, N= 100 people probability of at least say 40% of Michelob drinkers prefer Schlitz? Answer = P(40 Schlitz) + P(41 Schlitz)+.. + P(100 Schlitz) P (k Schlitz) = C(100,k)0.5 k 0.5 100 k REF: DAILYHIIT.COM.

MICHELOB VS. SCHLITZ When the sample size is large enough, the binomial distribution with parameters n and p can be approximated by the normal model with parameters

MICHELOB VS. SCHLITZ When the sample size is large enough, the binomial distribution with parameters n and p can be approximated by the normal model with parameters at least 40?

MICHELOB VS. SCHLITZ When the sample size is large enough, the binomial distribution with parameters n and p can be approximated by the normal model with parameters Schlitz at least 40?

MICHELOB VS. SCHLITZ When the sample size is large enough, the binomial distribution with parameters n and p can be approximated by the normal model with parameters Schlitz µ = 100 0.5 = 50 at least 40? = p 100 0.5 0.5) = 5

MICHELOB VS. SCHLITZ When the sample size is large enough, the binomial distribution with parameters n and p can be approximated by the normal model with parameters Schlitz µ = 100 0.5 = 50 at least 40? = p 100 0.5 0.5) = 5 Z = 40 50 5 = 2

MICHELOB VS. SCHLITZ When the sample size is large enough, the binomial distribution with parameters n and p can be approximated by the normal model with parameters Schlitz µ = 100 0.5 = 50 at least 40? = p 100 0.5 0.5) = 5 Z = 40 50 5 = 2 P(Schlitz>40)= P(Z>-2) = 0.978 (approximation) 0.982 (real)

MICHELOB VS. SCHLITZ Schlitz with N=100, p=0.7 (that is 0.7 times, Michelob drinkers will distinguish Michelob) µ =100 0.3 = 30 Z = 40 30 4.58 =2.18 = p 100 0.7 0.3 =4.58 P(Schlitz>40)=P(Z>2.18) =1.46%

MICHELOB VS. SCHLITZ Schlitz with N=100, p=0.7 (that is 0.7 times, Michelob drinkers will distinguish Michelob) np = 0.7x100=70, µ =100 0.3 = 30 Z = 40 30 4.58 =2.18 = p 100 0.7 0.3 =4.58 P(Schlitz>40)=P(Z>2.18) =1.46%

MICHELOB VS. SCHLITZ Schlitz with N=100, p=0.7 (that is 0.7 times, Michelob drinkers will distinguish Michelob) np = 0.7x100=70, n(1-p)=100x0.3=30 µ =100 0.3 = 30 Z = 40 30 4.58 =2.18 = p 100 0.7 0.3 =4.58 P(Schlitz>40)=P(Z>2.18) =1.46%

MICHELOB VS. SCHLITZ Schlitz with N=100, p=0.7 (that is 0.7 times, Michelob drinkers will distinguish Michelob) np = 0.7x100=70, both bigger than 10, n(1-p)=100x0.3=30 µ =100 0.3 = 30 Z = 40 30 4.58 =2.18 = p 100 0.7 0.3 =4.58 P(Schlitz>40)=P(Z>2.18) =1.46%

MICHELOB VS. SCHLITZ Schlitz with N=100, p=0.7 (that is 0.7 times, Michelob drinkers will distinguish Michelob) np = 0.7x100=70, n(1-p)=100x0.3=30 both bigger than 10, we can use normal approximation µ =100 0.3 = 30 Z = 40 30 4.58 =2.18 = p 100 0.7 0.3 =4.58 P(Schlitz>40)=P(Z>2.18) =1.46%

MICHELOB VS. SCHLITZ Schlitz with N=100, p=0.7 (that is 0.7 times, Michelob drinkers will distinguish Michelob) np = 0.7x100=70, n(1-p)=100x0.3=30 both bigger than 10, we can use normal approximation µ =100 0.3 = 30 Z = 40 30 4.58 =2.18 = p 100 0.7 0.3 =4.58 P(Schlitz>40)=P(Z>2.18) =1.46%

NORMAL APPROXIMATION

NORMAL APPROXIMATION A recent study found that Facebook users get more than they give. For example:

NORMAL APPROXIMATION A recent study found that Facebook users get more than they give. For example: 40% of Facebook users in our sample made a friend request, but 63% received at least one request

NORMAL APPROXIMATION A recent study found that Facebook users get more than they give. For example: 40% of Facebook users in our sample made a friend request, but 63% received at least one request Users in our sample pressed the like button next to friends content an average of 14 times, but had their content liked an average of 20 times

NORMAL APPROXIMATION A recent study found that Facebook users get more than they give. For example: 40% of Facebook users in our sample made a friend request, but 63% received at least one request Users in our sample pressed the like button next to friends content an average of 14 times, but had their content liked an average of 20 times Users sent 9 personal messages, but received 12 12% of users tagged a friend in a photo, but 35% were themselves tagged in a photo

NORMAL APPROXIMATION A recent study found that Facebook users get more than they give. For example: 40% of Facebook users in our sample made a friend request, but 63% received at least one request Users in our sample pressed the like button next to friends content an average of 14 times, but had their content liked an average of 20 times Users sent 9 personal messages, but received 12 12% of users tagged a friend in a photo, but 35% were themselves tagged in a photo Any guesses for how this pattern can be explained?