Answer:
a. -1.60377
b. 0.25451
c. 0.344
d. Option b) 78th
Explanation:
The number of pieces in a regular bag of Skittles is approximately normally distributed with a mean of 38.4 and a standard deviation of 2.12.
a)What is the z-score value of a randomly selected bag of Skittles that has 35 Skittles? a) 1.62 b) -1.62 c) 3.40 d) -3.40 e)1.303.
The formula for calculating a z-score is is z = (x-μ)/σ,
where x is the raw score
μ is the population mean
σ is the population standard deviation.
z = 35 - 38.4/2.12
= -1.60377
Option b) -1.62 is correct
b) What is the probability that a randomly selected bag of Skittles has at least 37 Skittles? a) .152 b) .247 c) .253 d).747e).7534. .
z = (x-μ)/σ
Mean of 38.4 and a standard deviation of 2.12.
z = (37 - 38.4)/2.12
= -0.66038
P-value from Z-Table:
P(x<37) = 0.25451
The probability that a randomly selected bag of Skittles has at least 37 Skittles is 0.25451
Option c) .253 is.correct
c) What is the probability that a randomly selected bag of Skittles has between 39 and 42 Skittles? a) .112 b) .232 c) .344 d).457 e).6125.
z = (x-μ)/σ
Mean of 38.4 and a standard deviation of 2.12.
For 39 Skittles
z = (39 - 38.4)/2.12
= 0.28302
Probability value from Z-Table:
P(x = 39) = 0.61142
For 42 Skittles
z = (42 - 38.4)/2.12
= 1.69811
Probability value from Z-Table:
P(x = 42) = 0.95526
The probability that a randomly selected bag of Skittles has between 39 and 42 Skittles is:
P(x = 42) - P(x = 39
0.95526 - 0.61142
0.34384
= 0.344
Option c is.correct
d) What is the percentile rank of a randomly selected bag of Skittles that has 40 Skittles in it? a)82nd b) 78th c) 75th d)25th e)22nd
z = (x-μ)/σ
Mean of 38.4 and a standard deviation of 2.12.
z = (40 - 38.4)/2.12
= 0.75472
P-value from Z-Table:
P(x = 40) = 0.77479
Converting to percentage = 0.77479× 100
= 77. 479%
≈ 77.5
Percentile rank = 78th