Demand per hour for gasoline at a local station is normally distributed with a mean of 1510 gallons and std deviation of 57 gallons. What is the probability that demand for a particular hour is greater than 1600 gallons? Enter your answer as a decimal, rounded to 4 decimal places

0.057168

Step-by-step explanation:

Given :

Mean = 1510

Standard deviation = 57

P(X > 1600)

To obtain the Zscore we use

(x - mean) / σ

P(Z > 1600) = P(Z > (x - mean) / σ)

P(Z > (x - mean) / σ) = P(Z > (1600 - 1510) / 57)

P(Z > 1.579) = 0.057168 (Z probability calculator)

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;

i would love   to but there is no graph shown

step-by-step explanation:

i am so sorry