A man who moves to a new city sees that there are two routes he could take to work. A neighbor who has lived there a long time he tells him Route A will average 5 minutes faster than Route B. The man decides to experiment. Each day he flips a coin to determine which way to go, driving each route 10 days. He finds that Route A takes an average of 49 minutes, with standard deviation 2 minutes, and Route B takes an average of 50 minutes, with standard deviation 4 minutes. Histograms of travel times for the routes are roughly symmetric and show no outliers. (alpha =0.05)
Find a 95% confidence interval for the difference between the Route B and Route A commuting times.

The 95% confidence interval for the difference between the Route B and Route A commuting times is -3.771808 to 1.771808

Step-by-step explanation:

Here we have the confidence interval is given by the following equation;

= 49 Minutes

σ₁ = 2 Minutes

= 50 Minutes

σ₂ = 4 Minutes

n₁ = n₂ = 10 days

α = 0.05

At 95%, from the z score table , z = 1.959964 ≈ ± 1.96

Plugging in the values Solving we get

Δμ Min = -3.771808 and Δμ Max = 1.771808.

answer is 67%

step-by-step explanation:

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.