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Use the information in [link] to answer the following questions.

  1. Find the 30 th percentile, and interpret it in a complete sentence.
  2. What is the probability that the age of a randomly selected smartphone user in the range 13 to 55+ is less than 27 years old.

Let X = a smart phone user whose age is 13 to 55+. X ~ N (36.9, 13.9)

  1. To find the 30 th percentile, find k such that P ( x < k ) = 0.30.
    invNorm(0.30, 36.9, 13.9) = 29.6 years
    Thirty percent of smartphone users 13 to 55+ are at most 29.6 years and 70% are at least 29.6 years.
  2. Find P ( x <27)
    This is a normal distribution curve. The peak of the curve coincides with the point 36.9 on the horizontal axis. The point 27 is also labeled. A vertical line extends from 27 to the curve. The area under the curve to the left of 27 is shaded. The shaded area shows that P(x < 27) = 0.2342.

    normalcdf(0,27,36.9,13.9) = 0.2342
    (Note that normalcdf(–10 99 ,27,36.9,13.9) = 0.2382. The two answers differ only by 0.0040.)
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There are approximately one billion smartphone users in the world today. In the United States the ages 13 to 55+ of smartphone users approximately follow a normal distribution with approximate mean and standard deviation of 36.9 years and 13.9 years respectively. Using this information, answer the following questions (round answers to one decimal place).

a. Calculate the interquartile range ( IQR ).

a.

  • IQR = Q 3 Q 1
  • Calculate Q 3 = 75 th percentile and Q 1 = 25 th percentile.
  • invNorm(0.75,36.9,13.9) = Q 3 = 46.2754
  • invNorm(0.25,36.9,13.9) = Q 1 = 27.5246
  • IQR = Q 3 Q 1 = 18.7508

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b. Forty percent of the ages that range from 13 to 55+ are at least what age?

b.

  • Find k where P ( x > k ) = 0.40 ("At least" translates to "greater than or equal to.")
  • 0.40 = the area to the right.
  • Area to the left = 1 – 0.40 = 0.60.
  • The area to the left of k = 0.60.
  • invNorm(0.60,36.9,13.9) = 40.4215.
  • k = 40.42.
  • Forty percent of the ages that range from 13 to 55+ are at least 40.42 years.

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Two thousand students took an exam. The scores on the exam have an approximate normal distribution with a mean μ = 81 points and standard deviation σ = 15 points.

  1. Calculate the first- and third-quartile scores for this exam.
  2. The middle 50% of the exam scores are between what two values?
  1. Q 1 = 25 th percentile = invNorm(0.25,81,15) = 70.9
    Q 3 = 75 th percentile = invNorm(0.75,81,15) = 91.9
  2. The middle 50% of the scores are between 70.9 and 91.1.
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A citrus farmer who grows mandarin oranges finds that the diameters of mandarin oranges harvested on his farm follow a normal distribution with a mean diameter of 5.85 cm and a standard deviation of 0.24 cm.

a. Find the probability that a randomly selected mandarin orange from this farm has a diameter larger than 6.0 cm. Sketch the graph.

a. normalcdf(6,10^99,5.85,0.24) = 0.2660

This is a normal distribution curve. The peak of the curve coincides with the point 2 on the horizontal axis. The values 1.8 and 2.75 are also labeled on the x-axis. Vertical lines extend from 1.8 and 2.75 to the curve. The area between the lines is shaded.
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b. The middle 20% of mandarin oranges from this farm have diameters between ______ and ______.

b.

  • 1 – 0.20 = 0.80
  • The tails of the graph of the normal distribution each have an area of 0.40.
  • Find k1 , the 40 th percentile, and k2 , the 60 th percentile (0.40 + 0.20 = 0.60).
  • k1 = invNorm(0.40,5.85,0.24) = 5.79 cm
  • k2 = invNorm(0.60,5.85,0.24) = 5.91 cm

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c. Find the 90 th percentile for the diameters of mandarin oranges, and interpret it in a complete sentence.

c. 6.16: Ninety percent of the diameter of the mandarin oranges is at most 6.15 cm.

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Using the information from [link] , answer the following:

  1. The middle 45% of mandarin oranges from this farm are between ______ and ______.
  2. Find the 16 th percentile and interpret it in a complete sentence.
  1. The middle area = 0.40, so each tail has an area of 0.30.

    1 – 0.40 = 0.60

    The tails of the graph of the normal distribution each have an area of 0.30.

    Find k1 , the 30 th percentile and k2 , the 70 th percentile (0.40 + 0.30 = 0.70).

    k1 = invNorm(0.30,5.85,0.24) = 5.72 cm

    k2 = invNorm(0.70,5.85,0.24) = 5.98 cm

  2. normalcdf(5,1099,5.85,0.24) = 0.9998
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Questions & Answers

binomial distribution tends to normal distribution
Murali Reply
if sample size n very large and probability tends to 0.5 if these both conditions are satisfied by binomial distribution it would tends to normal distribution
sravani
n tends to infinite i.e large Probability tends to 0 i.e indefinitely small. np = lamda
Anji
the above is poison to Bin
Anji
no of trails n tends to indefinitely large..i.e infine neither p nor q is very small Then bin tends to normal
Anji
if the death of of the snow is my yard is normally distributed with the m is equals to 2.5 and what is the probability that a randomly chosen location with have a no that between 2.25 and 2.76
Sakshi Reply
hey
Shubham
🤔
Iqra
hello
Sakshi
hii
Rushikesh
helow
why Statistics so hard
Mohd
ho geya solve
Sakshi
it's not hard
Sakshi
it is hard 😭
Mohd
solution?
Abdul
hii
Aadil
it's just need to be concentrate
Akinyemi
exactly..... concentration is very important
Iqra
rewrite the question
Aadil
what is the true statement about random variable?
Henna Reply
A consumer advocate agency wants to estimate the mean repair cost of a washing machine. the agency randomly selects 40 repair cost and find the mean to be $100.00.The standards deviation is $17.50. Construct a 90% confidence interval for the mean.
Deshah Reply
pls I need understand this statistics very will is giving me problem
Bolanle Reply
Sixty-four third year high school students were given a standardized reading comprehension test. The mean and standard deviation obtained were 52.27 and 8.24, respectively. Is the mean significantly different from the population mean of 50? Use the 5% level of significance.
Daryl Reply
No
Ariel
how do I find the modal class
Bruce Reply
look for the highest occuring number in the class
Kusi
the probability of an event occuring is defined as?
James Reply
The probability of an even occurring is expected event÷ event being cancelled or event occurring / event not occurring
Gokuna
what is simple bar chat
Toyin Reply
Simple Bar Chart is a Diagram which shows the data values in form of horizontal bars. It shows categories along y-axis and values along x-axis. The x-axis displays above the bars and y-axis displays on left of the bars with the bars extending to the right side according to their values.
Muhammad
statistics is percentage only
Moha Reply
the first word is chance for that we use percentages
muhammad
it is not at all that statistics is a percentage only
Shambhavi
I need more examples
Luwam Reply
how to calculate sample needed
Jim Reply
mole of sample/mole ratio or Va Vb
Gokuna
how to I solve for arithmetic mean
Joe Reply
Yeah. for you to say.
James
yes
niharu
how do I solve for arithmetic mean
Joe Reply
please answer these questions
niharu
add all the data and divide by the number of data sets. For example, if test scores were 70, 60, 70, 80 the total is 280 and the total data sets referred to as N is 4. Therfore the mean or arthritmatic average is 70. I hope this helps.
Jim
*Tan A - Tan B = sin(A-B)/CosA CosB ... *2sinQ/Cos 3Q = tan 3Q - tan Q
Ibraheem Reply

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Source:  OpenStax, Introductory statistics. OpenStax CNX. May 06, 2016 Download for free at http://legacy.cnx.org/content/col11562/1.18
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