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Question

the following data represents the following data represents the following data represents 2. Accordi ng to the Commerce Department in the Fall of 2014, 60% of US households had form of high-speed internet connection. Suppose 30 U.S. households were selected at random, find the probability that: a. Exactly 15 of these had a high-speed internet connection b. Between 16 and 20 (inclusive) had a high-speed C. Less than 17 had a high-speed internet connection. d. More than 19 had a high-speed internet connection internet connection. Suppose the ages of multiple-birth mothers (3 or more babies) are normally distribut mean age of 31.7 years and a standard deviation of 5.2 years. a. What percent of these mothers are between the ages 30 to 35. b. What percent of these mothers are less than 30 years old. c. What percent of these mothers are more than 33 years old 3. dda ocante tha umhar of dauc ahcent y and the final erade. v

Explanation / Answer

Question 2

Here number of households = 30

Pr(those who have high speed internet connection) = 0.60

(i) If x is the persons out of 30 who has good high speed internet connection

Pr(x = 15 ; 30 ; 0.60) = 30C15 (0.6)15 (0.4)15 = 0.0783

(ii) Pr(16   x    20) = BINOMCDF (x   20 ; 30 ; 0.60) - BINOMCDF (x < 16 ; 30 ; 0.60) = 0.8237 - 0.1754 = 0.6483

(iii) Pr(x < 17) = BINOMCDF (x < 17 ; 30 ; 0.60) = 0.2855

(iv) Pr(x > 19) = 1 - BINOMCDF(x   19 ; 30 ; 0.60) = 1 - 0.7085 = 0.2915

Question 3

Here mean age = 31.7 years

standard deviation = 5.2 years

(a) if x is the age of any random woman with three of more babies

Pr(30 < x < 35) = NORMCDF (x < 35 ; 31.7 ; 5.2) - NORMCDF(x < 30 ; 31.7 ; 5.2)

Z2 = (35 - 31.7)/5.2 = 0.6346

Z1 = (30 - 31.7)/5.2 = -0.3269

Pr(30 < x < 35) = Pr(Z < 0.6346) - Pr(Z < -0.3269) = 0.7372 - 0.3719 = 0.3653

(b) Pr(x < 30) = Pr(Z < -0.3269) = 0.3719

(c) Pr(x > 33) = 1 - Pr(Z < 33)

Z = (33 - 31.7)/5.2 = 0.25

Pr(x > 33) = 1 - Pr(Z < 33) = 1 - Pr(Z < 0.25) = 1 - 0.5987 = 0.4013

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