A large number of voltages show a mean of 118.5v and a population standard deviation of 1.20v. determine the percentage of data that falls between 116 and 120 volts.

Step-by-step explanation:

Hello!

X: voltage, it has a mean of μ= 118.5V and a standard deviation of σ= 1.20V.

Assuming the variable has a normal distribution, I'll use the standard normal distribution to calculate the probability that the voltage is between 116 and 120 volts.

P(116≤X≤120) = P(X≤120) - P(X≤116)

Using Z= (X - μ)/σ you now standardize the voltage values and get the cumulative probabilities from the tables of Z.

P(Z≤(120-118.5)/1.20) - P(Z≤(116-118.5)/1.20)

P(Z≤1.25) - P(Z≤-2.08)= 0.89435 - 0.01876= 0.87559

87.559% of the data will be between 116 and 120 volts.

I hope you have a SUPER day!

47.28 or 47.2857143

step-by-step explanation:

you just divide 331 by 7 and you get that answer. ( i think so tho)

answer:   35 % is the probability of a late departure or an early arrival.

step-by-step explanation:

let l represents the late departure and e represents the early departure.

then, according to the question,

p(l) = 12 %,   p(e) = 27 % and = 4 %

and, we have to find out,

but, we know that,

which is the required equation that will use to find the probability of a late departure or an early departure.

therefore,   the probability of a late departure or an early arrival =   35 %