Ajournal published a study of the lifestyles of visually impaired students. using diaries, the students kept track of several variables, including number of hours of sleep obtained in a typical day. these visually impaired students had a mean of 8.888.88 hours and a standard deviation of 2.092.09 hours. assume that the distribution of the number of hours of sleep for this group of students is approximately normal. complete parts a through c. a. find p(x < 6)b. find p(8< =x< =10)c. find the value for which p(x

a) P(X < 6) = 0.0838

b) P(8<=x<=10) = 0.3682

c) a = 7.1244

Step-by-step explanation:

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

a. Find P(x < 6)

This is the pvalue of Z when X = 6. So:

has a pvalue of 0.0838.

So P(X < 6) = 0.0838

b. Find P(8<=x<=10)

This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8. So:

X = 10

has a pvalue of 0.7054.

X = 8

has a pvalue of 0.3372.

So P(8<=x<=10) = 0.7054 - 0.3372 = 0.3682

c. Find the value for which P(x < a) = 0.2

This is X = a when Z has a pvalue of 0.2. So

7! = 7×6×5×4×3×2

3! = 3× 2

as 7! is multiple of 3! because 7! contains 3! that is 3× 2 as factor

so 7! /(3! ) = 7× 6× 5× 4

= 42 × 20

= 840

its b) 840✌✌✌

dhruv⭐