Assume that women's heights are normally distributed with a mean given by mu equals 62.5 in, and a standard deviation given by sigma equals 2.2 in.(a) if 1 woman is randomly selected, find the probability that her height is less than 63in.(b) if 37women are randomly selected, find the probability that they have a mean height less than 63in.(a)the probability is approximately nothing.(round to four decimal places as needed.)(b) the probability is approximately nothing.(round to four decimal places as needed.)

(a) 0.5899

(b) 0.9166

Step-by-step explanation:

Let X be the random variable that represents the height of a woman. Then, X is normally distributed with  

= 62.5 in

= 2.2 in

the normal probability density function is given by  

, then

(a) = 0.5899

   (in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2)

(b) We are seeking where n = 37. is normally distributed with mean 62.5 in and standard deviation . So, the probability density function is given by

, and

= 0.9166

(in the R statistical programming language) pnorm(63, mean = 62.5, sd = 2.2/sqrt(37))

You can use a table from a book to find the probabilities or a programming language like the R statistical programming language.

what statement? is there an image?