Question

The following data represent the weights​ (in grams) of a random sample of 50 candies.


0.87 0.88 0.82 0.90 0.90 0.84 0.84
0.91 0.94 0.86 0.86 0.86 0.88 0.87
0.89 0.91 0.86 0.87 0.93 0.88
0.83 0.95 0.87 0.93 0.91 0.85
0.91 0.91 0.86 0.89 0.87 0.84
0.88 0.88 0.89 0.79 0.82 0.83
0.90 0.88 0.84 0.93 0.81 0.90
0.88 0.92 0.85 0.84 0.84 0.86


Required:
a. Determine the sample standard deviation weight.
b. Use the Empirical Rule to determine the percentage of M&Ms with weights between 0.803 and 0.947 gram. Hint: = 0.875.
c. Determine the actual percentage of M&Ms that weigh between 0.803 and 0.947 gram, inclusive.
d. Use the Empirical Rule to determine the percentage of M&Ms with weights more than 0.911 gram.
e. Determine the actual percentage of M&Ms that weigh more than 0.911 gram.

Answer 1

Answer:

a)0.03558

b) 95%

c) 95.6985%

d) 16%

e)15.58%

Step-by-step explanation:

a) Determine the sample standard deviation weight.

Step 1

We find the Mean

Sum of terms/ Number of terms

Number of terms = 50

= 43.73/50

= 0.8746 grams

≈ 0.875 grams

Formula for sample standard deviation = √(x - Mean)²/n - 1

= √0.062042/50 - 1

= √0.001266163265

= 0.03558318796 grams

The empirical rule formula:

1) 68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

2) 95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

3)99.7% of data falls within 3 standard deviations from the mean - between μ - 3σ and μ + 3σ

b. Use the Empirical Rule to determine the percentage of M&Ms with weights between 0.803 and 0.947 gram. Hint: = 0.875.

95% of data falls within 2 standard deviations from the mean - between μ – 2σ and μ + 2σ .

μ – 2σ

0.875 - 0.03558 × 2

= 0.80384

μ + 2σ

= 0.875 + 0.03558 × 2

= 0.94616

Hence, 95% of M&Ms with weights between 0.803 and 0.947 gram.

c. Determine the actual percentage of M&Ms that weigh between 0.803 and 0.947 gram, inclusive.

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

For x = 0.803

= 0.803 - 0.875/0.03558

= -2.02361

Probability value from Z-Table:

P(x = 0.803) = 0.021505

For x = 0.947

= 0.947 - 0.875/0.03558

= = 2.02361

Probability value from Z-Table:

P(x = 0.947) = 0.97849

Hence,

P(x = 0.947 ) - P(x = 0.803)

= 0.97849 -0.021505

= 0.956985

Converting to percentage

= 0.956985 × 100

= 95.6985%

d. Use the Empirical Rule to determine the percentage of M&Ms with weights more than 0.911 gram.

68% of data falls within 1 standard deviation from the mean - that means between μ - σ and μ + σ .

=Mean = 0.875

Standard deviation = 0.03558

μ + σ

0.875 + 0.03558

= 0.91058

≈ 0.911

Hence, 68% of M&Ms with weights more than 0.911 gram.

Therefore, percentage of M&Ms with weights more than 0.911 gram, is on the right hand side(one side of the distribution)

=( 100 - 68)%/2

= 32%/2

= 16%

e. Determine the actual percentage of M&Ms that weigh more than 0.911 gram

The formula for calculating a z-score is is z = (x-μ)/σ, where x is the raw score, μ is the population mean, and σ is the population standard deviation.

Mean = 0.875

Standard deviation = 0.03558

x = 0.911

= 0.911 - 0.875/0.03558

= = 1.0118

Probability value from Z-Table:

P(x<0.911) = 0.84418

P(x>0.911) = 1 - P(x<0.911) = 0.15582

Converting to percentage =

0.15582 × 100

= 15.58%