Twelve residents from the city of Rocklin were randomly selected and asked "How many TVs are in your household?". The following data were obtained: 2, 3, 3, 1, 2, 5, 3, 4, 1, 2, 4, and 3. According to Nielsen Media Research the national average is 2.7 TVs per household. Is this sufficient sample evidence to indicate that the mean number of TVs in Rocklin households is higher than the reported national average of 2.7 TVs per household? Use ! = 5% and assume that the number of TVs in Rocklin households is normally distributed.

There isn't enough evidence to indicate that the mean number of TVs in Rocklin households is higher than the reported national average .

Step-by-step explanation:

This is a one sample mean test ;

H0 : μ = 2.7

H1 : μ > 2.7

Given the data :

2, 3, 3, 1, 2, 5, 3, 4, 1, 2, 4, 3

Sample size, n = 12

The sample mean, xbar = ΣX / n = 33/12 = 2.75

The sample standard deviation, s = 1.215 ( from calculator)

The test statistic :

(xbar - μ) ÷ (s/√(n))

T = (2.75 - 2.7) ÷ (1.215/√(12))

T = 0.05 / 0.3507402

T = 0.1426

The critical value from Tscore ;

df = 12 - 1 = 11

Critical value = 1.796

Since ; Test statistic < Critical value ;WE fail to reject the Null and conclude that there isn't enough evidence to indicate that the mean number of TVs in Rocklin households is higher than the reported national average

the correct option is 2.

step-by-step explanation:

according to the given information triangle abc is a right angle triangle with right angle at a. segment ab is 5 and segment ac is 7. point d is on segment bc and angles adb and adc are right angles.

it is given that triangles abd, cad, and cba are similar.

two triangles are similar if their corresponding sides are proportional.

in triangle cba, using pythagoras theorem

                      )

triangle abd and cba are similar,

                            (2)

triangle cad and cba are similar,

                      )

using (1), (2) and (3), we get

hence proved.

therefore option 2 is correct.

step-by-step explanation:

i found the life tables for the year 2012. 

life table of males at age 30. probability of dying between ages x and x+1

total population:  

29–30 . . . . . . . . . . . . 0.001030

30–31 . . . . . . . . . . . . 0.001061

males:  

29–30 . . . . . . . . . . . . 0.001420

30–31 . . . . . . . . . . . . 0.001448

white population

29–30 . . . . . . . . . . . . 0.001005

30–31 . . . . . . . . . . . . 0.001037

white males

29–30 . . . . . . . . . . . . 0.001372

30–31 . . . . . . . . . . . . 0.001399

non hispanic white population

29–30 . . . . . . . . . . . . 0.001083

30–31 . . . . . . . . . . . . 0.001116

non hispanic white males

29–30 . . . . . . . . . . . . 0.001477

30–31 . . . . . . . . . . . . 0.001503

jacob, non hispanic white male. 30 years old.

probability that jacob lives to be 30 years old. 

since the given probabilities are probabilities of dying, i think we simply subtract it from 1 to get the probability of living. we will use the probability of non hispanic white male.

                                          probability of living to be 30 y.o.

29 - 30 : 1 - 0.001477 = 0.998523

30 - 31 : 1 - 0.001503 = 0.998497