Suppose the driving distance to work for residents of a certain community has a mean of 21 miles and a standard deviation of 3.6 miles. what is the probability that an individual drives between 19 and 26 miles to work? assume that the distribution is normal

Step-by-step explanation:

We start by defining the random variable X.

X : ''The driving distance to work for residents of a certain community''

X can be modeled as a normal random variable.

X ~ N (μ,σ)

Where μ is the mean and σ is the standard deviation.

For this problem :

X ~ N (21,3.6)

Where the unit for the mean and the standard deviation is miles.

We are looking for

Given a normal random variable,we can subtract it the mean and then divide by the standard deviation in order to obtain a new random variable ''Z''.

Where ''Z'' is a new normal random variable.This is called standardizing.

Z ~ N (0,1)

The mean of Z is 0 and the standard deviation is 1.

We can find the  cumulative probability distribution of Z in any table.

Φ(a) given a certain value ''a''.

For the problem :

Φ(1.3888) - Φ(-0.5555) = 0.9177 - 0.2877 = 0.63

The probability that an individual drives between 19 and 26 miles to work is 0.63

d. 24+-0.114

step-by-step explanation:

the question requires us to calculate the approximate 95% confidence interval for the mean number of ounces of ketchup per bottle in the sample given the following statistics;

sample mean = 24

standard deviation, σ = 0.4

sample size , n = 49

the confidence interval for a population mean is calculated using the formula;

sample mean ± z-score*(σ/sqrt(n))

the z-score associated with a 95% confidence interval, from the standard normal tables, is;

1.96

substituting these values into the formula;

24 ± 1.96*(0.4/7)

= 24 ± 0.112

from the alternatives given, choice d is closest to the above expression;