Question

The life expectancy in the US is 75 with a standard deviation of 7 years. A random sample of 49 individuals is selected. What is the probability that the same mean will be larger than 77 years?

Answer 1

Answer:

0.02275

Step-by-step explanation:

We use this z score formula to solve for this question

z-score is is z = (x-μ)/σ/√n

where x is the raw score

μ is the population mean

σ is the population standard deviation

From the above question,

x = 77

μ = 75

σ = 7

number of samples = 49

Hence, the z score = 77 - 75/ (7/√49)

= 2/(7/7)

= 2/1

= 2

Hence, the z score = 2

Using the z table to find probability

P(x<77) = 0.97725

P(x>77) = 1 - P(x<77)

= 1 - 0.97725

= 0.02275

Therefore, the probability that the same mean will be larger than 77 years

is 0.02275