Chapter 7 Stats

Characteristics of the Normal Density Curve

REMEMBER THE PROPERTIES OF A BELL-SHAPED CURVE 1. The mean/median is located at the center of distribution, which is the highest point of the curve. 2. The points at x = µ - σ & x = µ + σ are inflection points (where the curve from the center dips down..think curves) 3. σ (std) determines the spread of the curve about the mean 4. area under curve = 1 5. the curve is symmetric (bell shaped) --> area to the left of the mean = 0.5, the area to the right of the mean = 0.5 6. probabilities have to be above x-axis (values need to be positive) 7. The Empirical Rule (68, 95, 99.7) & (34, 13.5, 2.35) & ofc (x = µ + σ, x = µ - σ and so forth)

z sub alpha

zₐ

Practice 1: Find the following probabilities (a) P(z > 2.03) (b) P(z ≤ -1.34 (c) P(0.27 ≤ z ≤ 2.15)

(a) = 0.0212 (b) = 0.9012 (c) = 0.3778

Example 3: Determine the area under the standard normal curve that lies between the following z-scores. (a) z = -2.55 and z = 2.55 (b) z = -3.03 and z = 1.98

(a) P(-2.55 < x < 2.55) = 0.9892 (b) P(-3.03 < x < 1.98) = 0.9749 check page 7

Example 1: Determine the area under the standard normal curve that lies to the left of the following. (a) z = -3.44 (b) z = 0.92

(a) P(x < -3.44) = 0.000291 --> check page 7 to see if you graphed correctly (b) P(x < 0.92) = 0.8212 --> check page 7 to see if you graphed correctly

Example 2: Determine the area under the standard normal curve that lies to the right of the following. (a) z = -0.55 (b) z = 2.23

(a) P(x > -0.55) = 0.7088 (b) P(x > 2.23) = 0.0129 check page 7

There are two normal density curves. One has a mean = 0 and std = 1. The other has a mean = 3 and std = 1. Describe what this would look like on a graph?

- both would have the same shape (same std), but the centers of the curves would be different (different mean). --> mean = 0 would be to the left of the mean = 3 by 3 units

Example 4: The lives of fridges are normally distributed with mean µ = 14 years and standard deviation σ = 2.5 years (a) Draw a normal curve with the parameters labeled. Then, shade the region that represents the proportion of fridges that last for more than 17 years.

- since mean and std are given, treat this like an empirical rule problem when doing the graph. --> HOWEVER when dealing with the probability we are dealing with intervals -----> WE CAN estimate the shaded area but cannot give the actual area!!!!! (check page 6 of chapter 7 notes for answer)

the smaller the std...

- the smaller the spread, and the higher the peak (because the with would be smaller)

There are two normal density curves. One has a mean = 0 and std = 1. The other has a mean = 0 and std = 2. Describe what this would look like on a graph?

- the two would be on top of each each, since they share the same mean. - but because the spread is different (the std is not the same), one would be flatter and more spread out (std = 2), while the (std = 1) would be tall and skinny.

Example 8: Scores on the Stanford-Binet Intelligence Test are normally distributed with mean µ = 100 and standard deviation σ = 16 (a) Determine the 90th percentile of IQs

AREA ON THE LEFT IS GIVEN TO US (90%) 90th percentile of IQs = 121

the left area of the z-score

1 - zₐ

Example 4: The lives of fridges are normally distributed with mean µ = 14 years and standard deviation σ = 2.5 years (b) Suppose that the area under the normal curve to the right of x = 17 is 0.1151. Provide two interpretations of this result.

1) 11.51% of fridges last more than 17 years 2) 0.1151 is the probability that a fridge would last more than 17 years

How to figure out the probability of a continuous random variable, WHEN the shape of the graph is uniform.

1) know that the total probability is 1 when dealing with P(x), which is relative frequency, and the random variable. 2) figure out the Length & Width of the rectangle on the graph. 3) Usually will be given the x (random variable), which represents the Length 4) L • W = 1 (the probability = area of rectangle) --> with this equation you can figure out the width. 5) Once you know both the width and length you can figure out the probability of what is being asked BY multiplying the length and width of the specified intervals.

unusual??

1) less than 0.05 (or 5%) 2) outside 2 std

Two Properties of Probability Density Function (pdf)

1. The total area of the shape in the graph (over all possible values of the random variable) must = 1 --> THINK RECTANGLE 2. The height of the shape on the graph must be greater than or equal to 0. BASICALLY, the SHAPE on the graph must lie above the x-axis (so all the probabilities must be positive!!!!)

how to round for z-scores

2 decimal places

Example 8: Scores on the Stanford-Binet Intelligence Test are normally distributed with mean µ = 100 and standard deviation σ = 16 (b) Determine the IQs that make up the middle 70% of IQs

70% of IQs is (83, 117)

The reaction time x (in minutes) of a certain chemical process follows a uniform probability distribution with 5 ≤ x ≤ 10. (a) Draw the graph of the density curve

Don't get muddle with the wording!! JUST DRAW A GRAPH OF PROBABILITY. P(x) ^ | | `------------` | | | | | | --------------------- > 5 10 reaction time (in minutes) - where the rectangle stops on the y-axis doesn't matter because the distribution is uniform!!! as long as the shape is a rectangle and matches up with the intervals that the variable lies within

We compute probabilities of continuous random variables over

INTERVALS of numbers unlike discreet random variables

normal density curve =

IT'S BASICALLY A BELL-SHAPED GRAPH --> considered having a normal distribution (or rather symmetrical)

Example 1: Your friend is always a minimum of 0 to 30 minutes late. Let the random variable X = the # of minutes late your friend will be, with all 1-minute intervals of time btwn x=0 and x=30 equally likely. The random variable X can be any value in the interval from 0 to 30, that is, 0 ≤ x ≤ 30. a) What is the probability that your friend will be between 5 and 10 minutes late?

L • W = 1 30 • W = 1 W = 1/30 intervals: 5 & 10 --> L = 5 prob = L • W = 5 • 1/30 = 1/6

The reaction time x (in minutes) of a certain chemical process follows a uniform probability distribution with 5 ≤ x ≤ 10. (c) What is the probability that the reaction time is less than 8 minutes?

L • W = 1 5 • W = 1 W = 1/5 intervals: 5-8 --> L = 3, W = 1/5 P = L • W = 3 • 1/5 = 3/5

The reaction time x (in minutes) of a certain chemical process follows a uniform probability distribution with 5 ≤ x ≤ 10. (b) What is the probability that the reaction time is between 6 and 8 minutes?

L • W = 1 5 • W = 1 W = 1/5 intervals: 6-8 --> L = 2, W = 1/5 P = L • W = 2 • 1/5 = 2/5

Example 7: The lengths of human pregnancies are approximately normally distributed with mean µ = 266 days and standard deviation σ = 16 days • What proportion of pregnancies lasts between 240 and 280 days?

P(240 < x < 280) = 0.7571

(c) Find the z-score that separates the middle 70% of the area under the standard normal curve from the area in the tales.

P(___ ≤ x ≤ ___) = .70 z = -1.0364 z = 1.0364 check page 9 for graph

(a) Find the z-score that the area under the standard normal curve to the left is 0.2

P(x < ___) = 0.2 z = -0.8416 check page 9 for graph

Example 9: The time required for Speedy Lube to complete an oil change service on a car approximately follows a normal distribution with mean 17 minutes and standard deviation of 2.5 minutes. If the service takes longer than 20 minutes, the customer receives it for half price. • What percent of customers receive it for half price?

P(x > 20) = 0.1151 = 11.51%

Example 7: The lengths of human pregnancies are approximately normally distributed with mean µ = 266 days and standard deviation σ = 16 days • What proportion of pregnancies lasts more than 270 days?

P(x > 270) = 0.4013 STAT CRUNCH mean: 266 std: 16 then P(x....

Example 9: The time required for Speedy Lube to complete an oil change service on a car approximately follows a normal distribution with mean 17 minutes and standard deviation of 2.5 minutes. If the service takes longer than 20 minutes, the customer receives it for half price. • If Speedy Lube doesn't want to give the discount to more than 3% of its customers, how long should it make the guaranteed time limit? Round to the nearest integer.

P(x > ___) = 0.03 guaranteed time limit: 22 minitues <-- answer

(b) Find the z-score that the area under the standard normal curve to the right is 0.35

P(x > ___) = 0.35 z = 0.3853 check page 9 for graph

Example 7: The lengths of human pregnancies are approximately normally distributed with mean µ = 266 days and standard deviation σ = 16 days • What is the probability that a randomly selected pregnancy lasts no more than 245 days?

P(x ≤ 245) = 0.095

Example 7: The lengths of human pregnancies are approximately normally distributed with mean µ = 266 days and standard deviation σ = 16 days • A "very preterm" baby is one whose gestation period is less than 224 days. Is this unusual?

TWO WAYS OF SOLVING unusual --> prob < 0.05 P(x < 224) = 0.004 (which is less than 0.05) -> yes unusual OR unusual --> outside 2 std z = value - mean / std = 224 - 266 / 16 = 2.63 2.63 --> outside 2 std -> yes, unusual

Standardizing a normal random variable

Z = x - μ /σ --> value - mean / std Z - score!!! but in this case we're using it for normal distribution.

Example 3: A random variable X is normally distributed with µ = 38.72 and σ = 3.17 (a) Compute Z1 = x1 - µ / σ for x1 = 35

Z1 = 35 - 38.72 / 3.17 = -1.17

Example 3: A random variable X is normally distributed with µ = 38.72 and σ = 3.17 (a) Compute Z2 = x2 - µ / σ for x2 = 38

Z2 = 38 - 38.72 / 3.17 = -0.23

Example 6: Find the value of zₐ a) z₀.₀₁ b) z₀.₀₂₅

a) P(x > __ ) = 0.01 z₀.₀₁ = 2.3263 check page 10 for graphs b) P(x > __ ) = 0.025 z₀.₀₂₅ = 1.96

Probability Density Function (pdf)

an equation used to compute probabilities of continuous random variables --> BASICALLY it's the name of the process

zₐ

the area to the right of the z-score --> the area would replace subscript a (so if the area was 1.45, you would right z₁.₄₅

what does it mean when we're given it's a standard normal curve?

mean = 0 and sd = 1

z distribution

mean = 0, sd = 1

when talking about the spread of data we're talking about

standard deviation LARGE std = spread is big amongst the data SMALL std = spread is small amongst the data

Example 4: Find the following probabilities (a) P(z < -0.61) (b) P(z ≥ -0.92) (c) P(-1.23 < z ≤ 1.61)

stat, calc, normal (a) = 0.2709 (b) = 0.8212 (c) = 0.8370

normal distribution on stat crunch

stat, calc, normal --> this is where we input n, p, the symbol, and either what the interval is or the probability (depending on what we're given)

z-score is

the interval on the curve

The curve on a graph is determined by (and how):

the mean and the standard deviation. --> the mean determines where the center of the curve is --> the standard deviation determines the spread of the curve (example: long and skinny, or fat and wide)

the area is

the probability

Since continuous random variables have infinite number of possible outcomes

the probability of observing a particular value is 0

A continuous random variable will have a graph of normal curve (or rather a bell-shaped distribution) if

the relative frequency histogram of the random variable is bell-shaped

if it's a uniform probability distribution...

the shape on the graph will be RECTANGLE

when dealing with standard normal curve, what happens with the symbols??

≤ and < mean the same thing in stat crunch (as does > ≥)