Question

Current research indicates that the distribution of the life expectancies of a certain protozoan is normal with a mean of 46 days and a standard deviation of 10.5 days. Find the probability that a simple random sample of 49 protozoa will have a mean life expectancy of 47 or more days.

Answer 1

Answer: P(x ≥ 47) = 0.25

Step-by-step explanation:

Since the distribution of the life expectancies of a certain protozoan is normal, we would apply the formula for normal distribution which is expressed as

z = (x - µ)/σ

Where

x = life expectancies of the certain protozoan.

µ = mean

σ = standard deviation

n = number of samples

From the information given,

µ = 46 days

σ = 10.5 days

n = 49

The probability that a simple random sample of 49 protozoa will have a mean life expectancy of 47 or more days is expressed as

P(x ≥ 47) = 1 - P(x < 47)

For x = 47

z = (47 - 46)/(10.5/√49) = 0.67

Looking at the normal distribution table, the probability corresponding to the z score is 0.0.75

P(x ≥ 47) = 1 - 0.75 = 0.25