A random sample of the price of gasoline from 40 gas stations in a region gives an average price of $3.69 and sample standard deviation 0.25$. Construct a 95% Confidence Interval for the mean price of regular gasoline in that region A. (3.61, 3.77) B. (3.09, 4.29) C. (2.79, 4.59) D. (3.39, 3.99)

95% Confidence Interval for the mean price of regular gasoline in that region

(3.61005, 3.76995)

Step-by-step explanation:

Step(i):-

Given sample size 'n' =40

mean of the sample x⁻ = 3.69 $

standard deviation of the sample "s' = 0.25 $

We will use students 't' distribution

degrees of freedom

ν = n-1 = 40-1 =39

Step(ii):-

95% Confidence Interval for the mean price of regular gasoline in that region

(3.69 - 0.07995 , 3.69 + 0.07995)

(3.61005, 3.76995)

Final answer:-

95% Confidence Interval for the mean price of regular gasoline in that region

(3.61005, 3.76995)

step-by-step explanation:

ok so im not 100% sure but you would start by adding 5.1t and 1.9t beacuse they have the same variable. that is 7t.   then add -3.6 + 1.2 which is -2.4

so then keep that as -2.4 + 7t. i apologize if this is incorrect. best luck on your test < 3

hello :

by applying the distributive property :

6 × 4.3 = 6 × ( 4 + 0.3 )

            =   (6 ×   4) + (6 ×   0.3)

            = 24 + 1.8

6 × 4.3 = 25.8