Question

A candy company claims that 15​% of its plain candies are​ orange, and a sample of 100 such candies is randomly selected.

a. Find the mean and standard deviation for the number of orange candies in such groups of 100.

μ=______​(Do not​ round.)

σ =______​(Round to one decimal place as​ needed.)

b. A random sample of 100 candies contains 16 orange candies. Is this result​ unusual? Does it seem that the claimed rate of 15​% is​ wrong?

A. Yes, because 16 is greater than the maximum usual value. ​Thus, the claimed rate of 15 % is not necessarily wrong.

B. ​Yes, because 16 is within the range of usual values.​ Thus, the claimed rate of 15 % is probably wrong.

C. Yes, because 16 is below the minimum usual value. ​Thus, the claimed rate of 15 % is not necessarily wrong.

D. No, because 16 is within the range of usual values. ​Thus, the claimed rate of 15 % is not necessarily wrong.

Answer 1

a)

Mean = n * p

= 100 * 0.15

= 15

Standard deviation = Sqrt ( np( 1 -p ) )

= Sqrt( 100 * 0.15 * 0.85)

= 3.6

b)

Minimum usual value = - 2

= 15 - 2 * 3.6

= 7.8

Maximum usual value = + 2

15 + 2 * 3.6

= 22.2

16 is within the range of the usual values.

No, because 16 is within the range of usual values. ​Thus, the claimed rate of 15% is not

necessarily wrong.