The time (in minutes) taken for a dose of a certain drug to be effective as a sedative on lab animals is normally distributed with mean =1 and variance 2=0.01. What is the proportion of animals for which the time taken is between 1 and 1.1 minutes?

Step-by-step explanation:

Let x be the random variable representing the time (in minutes) taken for a dose of a certain drug to be effective as a sedative on lab animals. Since it is normally distributed and the population mean and population standard deviation are known, we would apply the formula,

z = (x - µ)/σ

Where

x = sample mean

µ = population mean

σ = standard deviation

From the information given,

µ = 1

σ = √variance = √0.01 = 0.1

the probability that the time taken for a randomly selected animal is between 1 and 1.1 minutes is expressed as

P(1 ≤ x ≤ 1.1)

For x = 1,

z = (1 - 1)/0.1 = 0

Looking at the normal distribution table, the probability corresponding to the z score is 0.5

For x = 1.1

z = (1.1 - 1)/0.1 = 1

Looking at the normal distribution table, the probability corresponding to the z score is 0.84

Therefore,

P(1 ≤ x ≤ 1.1) = 0.84 - 0.5 = 0.34

The the proportion of animals for which the time taken is between 1 and 1.1 minutes is 0.34

1,150 give brainliest

step-by-step explanation

2r= 2300

r =1150

i got 142 as my answer

step-by-step explanation: