Normal Distribution. 6.4 Applications of Normal Distribution

Transcription

1 Normal Distribution 6.4 Applications of Normal Distribution 1 /20

2 Homework Read Sec 6-4. Discussion question p316 Do p316 probs 1-10, 16-22, 31, 32, 34-37, 39 2 /20

3 3 /20

4 Objective Find the probabilities for a normally distributed variable using the standard normal curve. 4 /20

5 5 /20 Objective: Students will find the probabilities for a normally Probability If a variable is normally distributed, or sufficiently unimodal and symmetrically distributed, we can use the standard normal curve to find the percentile of a score, the probability of a score or range of scores. We can also find a score corresponding to a probability or percentile. member?, you member! z = x µ σ

6 Percentile To find probability, we might convert an observation raw score into a z-score and use the table or calculator. We can also use the calculator to find the probability directly from the observed value if we know the mean and standard deviation of the distribution from which the observation is drawn. For instance, intelligence measured by the Wechsler Adult Intelligence Scale (WAIS) has µ = 100 and σ = 15. Thus a person with an IQ of 120 would have a z-score of z = = What does this tell us? 6/20

7 Graph Looking in the table, or using the calculator, we find that z-score corresponds to the percentile. z = = P(z ) = normalcdf(-99, , 0, 1) =.9088 OR P(x 120) = normalcdf(-10^99, 120, 100, 15) = Note the probability notation σ.9088 You must always draw the normal probability distribution and calculate the z-score when doing these problems. -3σ -2σ -1σ 0 1σ 2σ 3σ 7/20

8 Applications When finding solutions to these probability distribution questions there are four (4) components I will grade for at least six (6) points. 1. Draw a picture. Draw the Normal Distribution and show the appropriately shaded area in the normal probability density curve. 2. Calculate z-scores. Even if you use the raw scores to find the probability, you must calculate the appropriate z-scores. 3. Probability statement. You must write (a) the probability you calculated, (b) how it was found, and (c) the actual value. (e.g. (a) p(x 120) = (b) normalcdf(-999,120,100,15) = (c).9088 or p(z 1.33) = normalcdf(-9,1.33,0,1) = Answer the question. Answer the question in a complete sentence. 8/20

9 So? In a student body of 3100 students, how many students can we expect to have IQs above 120? σ P(z ) = = x 3100 = σ -2σ -1σ 0 1σ 2σ 3σ We can expect about 282 students out of 3100 will have an IQ of 120 or better. Remember: I insist, and will grade, you draw a graph (with some accuracy) to help you visualize the area of probability. 9/20

10 Gifted The term gifted is usually accepted to indicate 130 2σ an individual with an IQ 2σ above the mean This would equate to an IQ of σ -2σ -1σ 0 1σ 2σ 3σ Using the calculator, table, or remembering the empirical rule, this translates to about 2.28% of the population. P(x 130) = normalcdf(130, 999, 100, 15) = P(z 2) = normalcdf(2, 99, 0, 1) = In our population of 3100 students we can expect 3100(.02275) 70 (70.525) gifted students. 10/20

11 11 /20 Objective: Students will find the probabilities for a normally Example The average height of the adult U.S. male is 69.7 with σ = 2.9. Find the probability of an randomly selected adult male being between and 6 6 in height. -3σ -2σ -1σ 0 1σ 2σ 3σ z 1 = =.7931 z 2 = = P(.7931 z ) = Normalcdf(.7931, , 0, 1) =.2118 P(72 x 78) = Normalcdf(72, 78, 69.7, 2.9) =.2118 The probability of a randomly selected adult male is between 6 and 6 6 is.2118 Ensure the probability matches your calculation

12 12/20 Objective: Students will find the probabilities for a normally Example The average height of the adult U.S. female is 63.7 with σ = Find the probability of an randomly selected adult female being between and 6 6 in height. -3σ -2σ -1σ 0 1σ 2σ 3σ z 1 = = z 2 = = P(3.074 z ) = Normalcdf(3.0741, , 0, 1) =.0011 P(72 x 78) = Normalcdf(72, 78, 63.7, 2.7) =.0011 The probability of an adult female being between 6 and 6 6 is.0011

13 Example 25 At a music store the average purchase is $25 with σ = $ Find the probability that the next customer 40 spends between $25 and $ µ= $25 z = 0-3σ -2σ -1σ 0 1σ 2σ 3σ $40 z = = P(0 z 1.875) = Normalcdf(0, 1.875, 0, 1) =.4696 P(25 x 40) = Normalcdf(25, 40, 25, 8) =.4696 The probability that the next customer spends between $25 and $40 is /20

14 Example At a music store the average purchase is $25 with σ = $8 Find the probability that the next customer will spend at least $ σ -2σ -1σ 0 1σ 2σ 3σ P(z 1.875) = Normalcdf(1.875, 9, 0, 1) =.0304 P(x 40) = Normalcdf(40, 999, 25, 8) =.0304 The probability that the next customer will spend at least $40 is Find the probability that the next customer will spend less than $40. P(z 1.875) = Normalcdf(-9, 1.875, 0, 1) =.9696 P(x 40) Normalcdf(-999, 40, 25, 8) =.9696 The probability that the next customer will spend less than $40 is /20

15 15/20 Objective: Students will find the probabilities for a normally Finding Zemo We can also find the z-score associated with a known probability. P z=? We then convert the z-score back to an observed (raw) score by using x = zσ + µ

16 IQR for a Normal Model The weights of three year old females closely follows a Normal distribution with a mean of 30.7 pounds and a standard deviation of 3.6 pounds. Find the IQR for the distribution of the weights for three year old females. IQR = Q 3 - Q 2 (the middle 50%). First we will find the z scores bounding the middle 50%. z = invnorm(.25, 0, 1) = z = invnorm(.75, 0, 1) = We can find the x values bounding the middle 50% two ways z=? z= z=? z=.6745 x = zσ + µ x=? x=28.27 x=33.13 x=? x = (3.6) = pounds. x = (3.6) = pounds. z = invnorm(.25, 30.7, 3.6) = z = invnorm(.75, 30.7, 3.6) = The IQR is Q 3 - Q 2 = = 4.86 pounds. 16/20

17 Example A pizza restaurant delivers in a mean time of 18.2 minutes, with a standard deviation of 3 minutes. The company wants to guarantee delivery time or the pizza is free. If the company is willing to give away no more than 5%.05 of its pizza deliveries (longer delivery time), what would be the time to deliver they should guarantee? z= z=? P(z z x ) =.05 t= First we find z. z = invnorm(.95, 0, 1) = Then we convert z to the raw score. x = = (3) = min. Or we could just go straight to the raw score. P(t x) =.05 x = invnorm(.95, 18.2, 3) = minutes The company should guarantee delivery in 24 minutes. 17/20

18 18/20 Objective: Students will find the probabilities for a normally The pizza restaurant delivers in a mean time of 18.2 minutes and wants to guarantee a delivery time of 20 minutes or the pizza is free. That would require improving the consistency of delivery times. If the company is willing to give away no more than 5% of its pizza deliveries (longer delivery time), what standard deviation must they obtain?.05 First we find z. z = invnorm(.95, 0, 1) = Next we find s by using that z = s s = = The company can guarantee delivery in 20 minutes, giving away no more than 5% of the pizzas if the standard deviation is reduced to 1.09 minutes.

19 Z-score If you choose to go straight to the actual value you must still calculate the z-score. 19/20

20 Is the data Normal? Using z-scores requires that data be normally, or sufficiently normally, distributed. The best test for normality is drawing a histogram to see the shape of the distribution. If the data looks about normal we are happy. This histogram is positively skewed, but sufficiently symmetric for purposes of using the normal model. 20/20