Question

The average playing time of compact discs in a large collection is 37 min, and the standard deviation is 2 min.
(a) What value is 1 standard deviation above the mean? 1 standard deviation below the mean? What values are 2 standard deviations away from the mean?
1 standard deviation above the mean
1 standard deviation below the mean
2 standard deviation above the mean
2 standard deviation below the mean

(b) Without assuming anything about the distribution of times, at least what percentage of the times are between 33 and 41 min? (Round the answer to the nearest whole number.)
At least %

(c) Without assuming anything about the distribution of times, what can be said about the percentage of times that are either less than 31 min or greater than 43 min? (Round the answer to the nearest whole number.)

(d) Assuming that the distribution of times is normal, approximately what percentage of times are between 33 and 41 min? (Round the answers to two decimal places, if needed.)
%
Less than 31 min or greater than 43 min?
%
Less than 31 min?
%

Answer 1

A) 1 standard deviation above the mean = 37 + 2 = 39

1 standard deviation below the mean = 37 - 2 = 35

2 standard deviation above the mean = 37 + 2 * 2 = 41

2 standard deviation below the mean = 37 - 2 * 2 = 33

B) P(33 < X < 41) = 1 - 1/k²

= 1 - 1/2²

= 1 - 1/4

= 3/4 = 0.75 = 75%

C) P(X < 31) + P(X > 43)

= 1 - P(31 < X < 43)

= 1 - (1 - 1/k²)

= 1 - (1 - 1/3²)

= 1 - 1 + 1/9

= 0.1111 = 11.11%

D) P(33 < X < 41)

= P((33 - )/ < (X -  )/ < (41 -  )/)

= P((33 - 37)/2 < Z < (41 - 37)/2)

= P(-2 < Z < 2)

= P(Z < 2) - P(Z < -2)

= 0.9772 - 0.0228

= 0.9544 = 95.44%

P(X < 31) + P(X > 43)

= P((X -  )/ < (31 -  )/) + P((X - )/ > (43- ​​​​​ )/)

= P(Z < (31 - 37)/2/+ P(Z > (43 - 37)/2)

= P(Z < -3) + P(Z > 3)

= 0.0013 + (1 - 0.9987)

= 0.0013 + 0.0013

= 0.0026 = 0.26%

P(X < 31)

= P((X -  )/ < (31 - )/)

= P(Z < (31 - 37)/2)

= P(Z < -3)

= 0.0013 = 0.23©