Essential Statistics
Essential Statistics
2nd Edition
ISBN: 9781259570643
Author: Navidi
Publisher: MCG
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Chapter 5.1, Problem 46E

a.

To determine

Find the value of P(2).

a.

Expert Solution
Check Mark

Answer to Problem 46E

The value of P(2) is 0.258.

Explanation of Solution

Calculation:

The table represents the probability distribution of the random variable X, the number of teenagers who have sent text messages on their cell phones within past 30 days.

From the probability distribution table, the probability at the point x=2 is, P(1=2)=0.258.

Thus, the value of P(2) is 0.258.

b.

To determine

Find the probability value, P(More than 1).

b.

Expert Solution
Check Mark

Answer to Problem 46E

The probability value, P(More than 1) is 0.888.

Explanation of Solution

Calculation:

The required probability can be obtained as shown below:

P(More than 1)=P(x>1)=1P(x1)=1[P(x=0)+P(x=1)]

Substituting the value from the probability distribution table,

P(More than 1)=1[P(x=0)+P(x=1)]=1[0.015+0.097]=10.112=0.888

Thus, the probability value, P(More than 1) is 0.888.

c.

To determine

Find the probability that three or more of the teenagers sent text messages.

c.

Expert Solution
Check Mark

Answer to Problem 46E

The probability that three or more of the teenagers sent text messages is 0.63.

Explanation of Solution

Calculation:

The probability that three or more of the teenagers sent text messages is the sum of the probabilities at x=3,x=4 and x=5.

Substituting the values from the probability distribution table,

P(Three or more sent text messages)=P(x=3)+P(x=4)+P(x=5)=0.343+0.227+0.06=0.63

Thus, the probability that three or more of the teenagers sent text messages is 0.63.

d.

To determine

Find the probability that fewer than two of the teenagers sent text messages.

d.

Expert Solution
Check Mark

Answer to Problem 46E

The probability that fewer than two of the teenagers sent text messages is 0.112.

Explanation of Solution

Calculation:

The probability that fewer than two of the teenagers sent text messages is the sum of the probabilities at x=3 and x=4.

P(Fewer than two sent text messages)=P(x<2)=P(x=0)+P(x=1)

Substituting the values from the probability distribution table,

P(Fewer than two sent text messages)=P(x<2)=P(x=0)+P(x=1)=0.015+0.097=0.112

Thus, the probability that fewer than two of the teenagers sent text messages is 0.112.

e.

To determine

Find the mean.

e.

Expert Solution
Check Mark

Answer to Problem 46E

The mean value is 2.85.

Explanation of Solution

Calculation:

The formula for the mean of a discrete random variable is,

E(X)=μX=xP(x)

The mean of the random variable is obtained as given below:

xP(x)xP(x)
00.0150.000
10.0970.097
20.2580.516
30.3431.029
40.2270.908
50.0600.300
Total1.0002.850

Thus, the mean value is 2.85.

f.

To determine

Find the standard deviation.

f.

Expert Solution
Check Mark

Answer to Problem 46E

The standard deviation is 1.107.

Explanation of Solution

Calculation:

The standard deviation of the random variable X is obtained by taking the square root of variance.

The formula for the variance of the discrete random variable X is,

σX2=[(xμX)2P(x)]

Where μX=[xP(x)] represents the mean of the random variable X.

The variance of the random variable X is obtained using the following table:

xP(x)(xμX)(xμX)2(xμX)2P(x)
00.015–2.858.12250.122
10.097–1.853.42250.332
20.258–0.850.72250.186
30.3430.150.02250.008
40.2271.151.32250.300
50.0602.154.62250.277
Total1.000–2.118.2351.226

Therefore,

σ2=1.226

Thus, the variance is 1.226.

The standard deviation is,

σ=1.226=1.107

That is, the standard deviation is 1.107.

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Chapter 5 Solutions

Essential Statistics

Ch. 5.1 - Prob. 11ECh. 5.1 - Prob. 12ECh. 5.1 - Prob. 13ECh. 5.1 - Prob. 14ECh. 5.1 - Prob. 15ECh. 5.1 - Prob. 16ECh. 5.1 - Prob. 17ECh. 5.1 - In Exercises 17–26, determine whether the random...Ch. 5.1 - Prob. 19ECh. 5.1 - In Exercises 17–26, determine whether the random...Ch. 5.1 - Prob. 21ECh. 5.1 - In Exercises 17–26, determine whether the random...Ch. 5.1 - Prob. 23ECh. 5.1 - Prob. 24ECh. 5.1 - Prob. 25ECh. 5.1 - Prob. 26ECh. 5.1 - In Exercises 27–32, determine whether the table...Ch. 5.1 - In Exercises 27–32, determine whether the table...Ch. 5.1 - Prob. 29ECh. 5.1 - In Exercises 27–32, determine whether the table...Ch. 5.1 - Prob. 31ECh. 5.1 - In Exercises 27–32, determine whether the table...Ch. 5.1 - Prob. 33ECh. 5.1 - In Exercises 33–38, compute the mean and standard...Ch. 5.1 - Prob. 35ECh. 5.1 - In Exercises 33–38, compute the mean and standard...Ch. 5.1 - In Exercises 33–38, compute the mean and standard...Ch. 5.1 - In Exercises 33–38, compute the mean and standard...Ch. 5.1 - Prob. 39ECh. 5.1 - 40. Fill in the missing value so that the...Ch. 5.1 - 41. Put some air in your tires: Let X represent...Ch. 5.1 - Prob. 42ECh. 5.1 - Prob. 43ECh. 5.1 - Prob. 44ECh. 5.1 - Prob. 45ECh. 5.1 - Prob. 46ECh. 5.1 - Prob. 47ECh. 5.1 - 48. Pain: The General Social Survey asked 827...Ch. 5.1 - Prob. 49ECh. 5.1 - Prob. 50ECh. 5.1 - 51. Lottery: In the New York State Numbers...Ch. 5.1 - 52. Lottery: In the New York State Numbers...Ch. 5.1 - Prob. 53ECh. 5.1 - Prob. 54ECh. 5.1 - Prob. 55ECh. 5.1 - Prob. 56ECh. 5.1 - Prob. 57ECh. 5.1 - Prob. 58ECh. 5.1 - Prob. 59ECh. 5.1 - Prob. 60ECh. 5.1 - Prob. 61ECh. 5.2 - 1. Determine whether X is a binomial random...Ch. 5.2 - Prob. 2CYUCh. 5.2 - Prob. 3CYUCh. 5.2 - Prob. 4CYUCh. 5.2 - Prob. 5ECh. 5.2 - In Exercises 5–7, fill in each blank with the...Ch. 5.2 - Prob. 7ECh. 5.2 - In Exercises 8–10, determine whether the statement...Ch. 5.2 - Prob. 9ECh. 5.2 - In Exercises 8–10, determine whether the statement...Ch. 5.2 - Prob. 11ECh. 5.2 - Prob. 12ECh. 5.2 - Prob. 13ECh. 5.2 - In Exercises 11–16, determine whether the random...Ch. 5.2 - Prob. 15ECh. 5.2 - In Exercises 11–16, determine whether the random...Ch. 5.2 - Prob. 17ECh. 5.2 - In Exercises 17–26, determine the indicated...Ch. 5.2 - Prob. 19ECh. 5.2 - In Exercises 17–26, determine the indicated...Ch. 5.2 - In Exercises 17–26, determine the indicated...Ch. 5.2 - Prob. 22ECh. 5.2 - Prob. 23ECh. 5.2 - In Exercises 17–26, determine the indicated...Ch. 5.2 - Prob. 25ECh. 5.2 - Prob. 26ECh. 5.2 - Prob. 27ECh. 5.2 - 28. Take another guess: A student takes a...Ch. 5.2 - Prob. 29ECh. 5.2 - Prob. 30ECh. 5.2 - Prob. 31ECh. 5.2 - 32. What should I buy? A study conducted by the...Ch. 5.2 - Prob. 33ECh. 5.2 - Prob. 34ECh. 5.2 - Prob. 35ECh. 5.2 - Prob. 36ECh. 5.2 - Prob. 37ECh. 5.2 - 38. Stress at work: In a poll conducted by the...Ch. 5.2 - Prob. 39ECh. 5.2 - Prob. 40ECh. 5.2 - Prob. 41ECh. 5 - Prob. 1CQCh. 5 - Prob. 2CQCh. 5 - Prob. 3CQCh. 5 - Prob. 4CQCh. 5 - Prob. 5CQCh. 5 - Prob. 6CQCh. 5 - Prob. 7CQCh. 5 - Prob. 8CQCh. 5 - 9. At a cell phone battery plant, 5% of cell phone...Ch. 5 - Prob. 10CQCh. 5 - Prob. 11CQCh. 5 - Prob. 12CQCh. 5 - Prob. 13CQCh. 5 - Prob. 14CQCh. 5 - Prob. 15CQCh. 5 - Prob. 1RECh. 5 - Prob. 2RECh. 5 - Prob. 3RECh. 5 - Prob. 4RECh. 5 - Prob. 5RECh. 5 - 6. Lottery tickets: Refer to Exercise 5. What is...Ch. 5 - Prob. 7RECh. 5 - Prob. 8RECh. 5 - Prob. 9RECh. 5 - Prob. 10RECh. 5 - Prob. 11RECh. 5 - Prob. 12RECh. 5 - Prob. 13RECh. 5 - Prob. 14RECh. 5 - Prob. 15RECh. 5 - Prob. 1WAICh. 5 - Prob. 2WAICh. 5 - Prob. 3WAICh. 5 - Prob. 4WAICh. 5 - Prob. 5WAICh. 5 - Prob. 6WAICh. 5 - One of the most surprising probability...
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