NORMAL DISTRIBUTION Flashcards
symmetric distribution where the data values are evenly distributed about
mean
what is the shape of the normal distribution
Bell-shaped
It is the other term for the normal distribution
Gaussian Distribution
it is an important assumption in parametric test
Normal Distribution
In the normal distribution Mean, median, and mode are essentially___________
the same
not symmetric and extends more to one side than the other
Skewed
Left skewed is aka
negatively skewed
order of mean, median, mode in left- skewed
mean
order of mean, median, mode in right- skewed
mean>median>mode
A normal distribution can have any ______and any______________
mean and positive standard
deviation.
describes the spread of the data.
standard deviation
The higher the
standard deviation, the ________ the values are from
the mean.
more it is spread or the farther
if the __________ is increased the spread of
the data will increase also.
standard deviation
relationship of mean, median, and mode in normal distribution
they are equal
The normal curve is
shaped like a ________and symmetric about the_______
bell; mean.
The total area under the curve is equal to__
1
The normal curve approaches, but never touches the________as it extends
farther and farther away from the ________(Asymptotic)
x-axis; mean.
the number of standard deviations that a given x value is above or below
the mean
zScore
Non-standard /General Distribution
Left curve
Standard Normal Distribution
Right Curve
Standard Normal Distribution - the Mean is always __ and
standard deviation is always _____. Regardless the range of X values.
zero; one
P(X) and P(Z) corresponds to the area under the curve which is ____
one (1)
range of the Probability for a given value: _______________
0 ≤ P (Z) ≤ 1
Total percentage of the values is____ Positive infinity to negative
infinity is ____
100%
Empirical Rule: 1 standard deviation=___
68%
Empirical Rule: 2 standard deviation=___
95%
Empirical Rule: 3 standard deviation=___
99.7%
Calculating the z score given the area or probability.
(Reverse lookup
process)
First thing to do in finding the area or probability given an x value.
Convert x to z: Standardized first using the formula: z = x - µ / σ
find P (z≤ - 1.61) in percentage
5.37%
find P(z≤1.5) in percentage
93.32%
P (z ≤ -1.56)
0.0594
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of 5. Find the probability that a randomly selected student scored less than 65 on the exam.... what is the z-score?
0.4
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of 5. Find the probability that a randomly selected student scored less than 65 on the exam.
µ= 63; x = 65. σ =5 • P (x ≤ 65) • z=x- µ/σ = 65-63/5 = 0.4 • P (z ≤ 0.4) = 0.6554 • Ans: The probability is 65.54%
What is the probability of getting a standard normal value greater than
0.93?
P (z ≥ 0.93) = 0.8238 (area not shaded)
1-0.8238= 0.1762 (area shaded: looking for)
Ans: 17.62%
Rule for probability of complements
1 - P (z ≤ z)
The final exam scores in a statistics class were normally distributed
with a mean of 63 and a standard deviation of 5. Find the probability that a
randomly selected student scored more than 57 on the exam.
• The question is the blue shaded part
- The question is the blue shaded part.
- µ=63; x=57. σ =5
- P(x57): convert x to z
- z=x- µ/σ =57-63/5 =-1.2
- P (z ≤ -1.2) = 0.1151
- 1- P (z ≤ -1.2) = 1-0.1151 = 0.8849.
- Ans: the probability is 88.49%
Find the area under the standard normal curve between z =- 1.98 and
z= 1.07
• P (-1.98 ≤ z ≤ 1.07) • P (z ≤ 1.07) = 0.8577 • P (z ≥ -1.98)= 0.0239 • Area of the shaded part: 0.8577 - 0.0239 = 0.8338 → Ans: 0.8338
The systolic blood pressure (SBP) of adult men is normally distributed
with mean of 110 mmHg and standard deviation of 15 mmHg. How many
percent of adult men have systolic blood pressures between 90 mmHg and 125
mmHg?
• µ=110; 90 ≤ x ≤ 125; =15 • 1st: convert x to z → P (x ≤ 125) = 125-110/ 15 = 1 → P (x ≥ 90) = 90-110/ 15 = -1.33 • Z-score: → P (z ≤ 1) = 0.8413 → P (z ≥ -1.33) = 0.0918 • Area between the 2 z-scores: → 0.8413-0.0918= 0.7495 → ANS: 74.95%
The final exam scores in a statistics class were normally distributed with a mean of 63 and a standard deviation of 5. Find the probability that a randomly selected student scored between 55 and 70.
• µ=63; 55 ≤ x ≤ 70; =5 • 1st: convert x to z → P (x ≤ 70) = 70-63/ 5 = 1.4 → P (x ≥ 55) = 55-63/ 5 = -1.6 • Z-score: → P (z ≤ 1.4) = 0.9192 → P (z ≥ -1.6) = 0.0548 • Area between the 2 z-scores: 0.9192-0.0548= 0.8644 → ANS: 86.44%
Find the value of z1 in the figure below. The area to the left of z1 is 0.0485 or
P(z
Find the value of z2= 0.9224
Find the z-score above which lies 20% of scores in the normal distribution
→ Want to find the z-score that cuts the distribution into lower 80% and
upper 20%
→ Same as 80th percentile
→ Closest to the given area 0.8 is 0.7995
→ ANS: 0.84
Find the z-score that corresponds to P75.
Ask about the z score that cuts the distribution to lower 75% and upper
25%.
→ Area= 0.75; P75: lower 75%
→ 0.75: z-score = closest 0.7486 which is 0.67
Find x1 such that P (X
→ 0.9332 in z-score is 1.5
→ x=10+1.5(2.5) = 13.75
→ 1.5 above the mean
Find x1 such that P (X >x1) = 0.65, where X is a normal random variable with
μ = 175 and σ = 12
→ Determine the area to the left: 1- P(xx1) = 1- 0.65 = 0.35
→ 0.35 in z-score is -0.39 (.3483)
→ x=175+(-.39) (12) =170.32
→ x1=170.32 and located -.39 below the mean.
X is a normally distributed random variable with mean of 15 and standard
deviation of 0.25. Find the values x1 and x2 that are symmetrically located
with respect to the mean of X and satisfy P(x1
→ Areas to the right and to left of the shaded area have a value of 0.1
▪ Determine the z2:
→ Area = 0.9 -> (0.8: shaded area plus 0.1: left side of the shaded area)
▪ Determine the z1:
→ Area= 0.1 the left side area
▪ In the z-table: z2 =1.28 (0.8997)
▪ z1=-1.28 (0.1003)
Symmetric → x2, =15+1.28(0.25) =15.32 → x1, =15+(-1.28) (0.25) =14.68 → x1=14.68; x2=15.32 and x2 located 1.28 above the mean while x1 located -1.28 below the mean
