Question

i need help with with c and d. im using the T1-84 calculator for my class. thank you

X23. The Empirical Rule The following data represent the weights (in grams) of a random sample of 50 M&M plain 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 Source: Michael Sullivan (a) Determine the sample standard deviation weight. Express your answer rounded to three decimal places. (b) On the basis of the histogram drawn in Section 3.1, Problem 25, comment on the appropriateness of using the Empirical Rule to make any general statements about the weights of M&Ms. (c) Use the Empirical Rule to determine the percentage of M&Ms with weights between 0.803 and 0.947 gram. Hint: x = 0.875. X (d) Determine the actual percentage of M&Ms that weigh between 0.803 and 0.947 gram, inclusive. X(e) Use the Empirical Rule to determine the percentage of M&Ms with weights more than 0.911 gram. Determine the actual percentage of M&Ms that weigh more than 0.911 gram. data represent the length

Answer 1

Solution:

We are given data of weight of 50 M&M plain candies.

Part c) Using Empirical rule, we have to find the percentage of M&M's with weights between 0.803 and 0.947 gram.

We need to find standard deviation s.

Use TI-84 plus calculator to find sample standard deviation s.

Press STAT and select EDIT

Under Select column L1 and enter all 50 numbers

Then again press STAT and select CALC

Under CALC , select 1-Var Stats

Thus we get:

TI-84 Plus Silver Edition TEXAS INSTRUMENTS 1-Var Stats x=.8746 x=43.73 In=50 STAT PLOTF1 TBLSETE2 FORMAT E WINDOW ZOOM CALC F4 TRACE TABLE F5 GRAPH 2ND MODE DEL ALPHA [email protected] STAT MATH CLEAR APPS PRGM VARS

T = 0.8746

T = 0.875

and

Sy = 0.035583

S = 0.036

Now use given values to find number of standard deviations k.

Lower limit = 0.803 and Upper limit = 0.947

I - K x Sp = 0.803

0.875 - k x 0,036 = 0.803

0.875 – 0.803 = kx 0,036

0,072 = x 0,036

0.072 0,036 k x 0.036 0.036

that is:

and

$\bar{x} + k \times s_{x} = 0.947$

$0.875 + k \times 0.036 = 0.947$

k x 0.036 = 0.947 – 0.875

$k \times 0.036 = 0.072$

10.072 0.036

Thus values 0.803 and 0.947 are 2 standard deviation from mean.

Empirical rule:

1) 68% of the data falls within 1 standard deviation from mean

$P( \mu - \sigma < X < \mu + \sigma ) = 68 \%$

2) 95% of the data falls within 2 standard deviation from mean

$P( \mu - 2\sigma < X < \mu + 2 \sigma ) = 95 \%$

3) 99.7% of the data falls within 3 standard deviation from mean

$P( \mu - 3\sigma < X < \mu + 3 \sigma ) = 99.7 \%$

Thus according to empirical rule: 95% of the data falls within 0.803 and 0.947 gram.

Part d) Actual percentage of M&M's that weigh between 0.803 and 0.947 gram inclusive.

From given data set , we have only one value = 0.79 < 0.803 and only one value = 0.95 > 0.947 ,

thus we have = 50-2 = 48 M&Ms candies that weigh between 0.803 and 0.947 gram inclusive

Thus percentage of M&M's that weigh between 0.803 and 0.947 gram inclusive = 48 / 50 = 0.96

Thus percentage of M&M's that weigh between 0.803 and 0.947 gram inclusive = 96%