Butter side down Refer to the preceding exercise. Maria decides to test this

probability and drops 10 pieces of toast from a 2.5-foot table. Only 4of them land butter

side down. Maria wants to perform a simulation to estimate the probability that 4or

fewer pieces of toast out of 10would land butter side down if the researchers’ 0.81

probability value is correct.

a. Describe how you would use a table of random digits to perform the simulation.

b. Perform 3trials of the simulation using the random digits given. Copy the digits onto

your paper and mark directly on or above them so that someone can follow what you

did.

29077
14863
61683
47052
62224
51025
95052
90908
73592
75186
87136
95761
27102
56027
55892
33063
41842
81868

c. The dotplot displays the results of 50 simulated trials of dropping 10pieces of toast.

Is there convincing evidence that the researchers’ 0.81probability value is incorrect?

Explain your answer.

Short Answer

Expert verified

(a)First we will select a row and then we will select a two - digit number,

if the number lies between 00to 80(both including),

then toast will land butter side down,otherwise

toast will land butter side up.

(b)First trial:10toast lie butter side down and 0toast lie butter side up.

Second trial:9toast lie butter side down and 1 toast lie butter side up.

Third trial: 8toast lie butter side down and 2 toast lie butter side up.

(c)No, there is no convincing evidence that the researchers’ 0.81probability value is incorrect.

Step by step solution

01

Part (a) Step 1:Given Information

We have been given that,

Maria wants to perform a simulation , the probability that 4or fewer pieces of toast out of 10would land butter side down if the researchers’ probability value is correct =0.81

probability of toast landing butter side down =0.81

29077
14863
61683
47052
62224
51025
95052
90908
73592
75186
87136
95761
27102
56027
55892
33063
41842
81868
02

Part (a) Step 2:Explanation

First we will select a row and then we will select a two - digit number,

if the number lies between 00to 80(both including),

then toast will land butter side down,otherwise

toast will land butter side up.

Number of trials given to perform = 10

So,we will select 10numbers.

Record the number of toast that land butter side down out of 10.

03

Part (b) Step 1:Given Information

We have been given that,

Maria wants to perform a simulation , the probability that 4or fewer pieces of toast out of 4would land butter side down. if the researchers’ probability value is correct =0.81

probability of toast landing butter side down =0.81

04

Part (b) Step 2:Explanation

First we will select a row and then we will select a two - digit number,

if the number lies between 00to 80(both including),

then toast will land butter side down,otherwise

toast will land butter side up.

We have selected first row and the first two-digit number selected is 29.

First trial:

numbers selected are,

29,07,71,48,63,61,68,34,70,52

all the selected numbers are between 00to 80.

So,10toast lie butter side down and 0toast lie butter side up.

Second trial:

numbers selected are,

62,22,45,10,25,95,05,29,09,08

all the selected numbers are between 00to 80except 95.

So, 9toast lie butter side down and 1toast lie butter side up.

Third trial:

numbers selected are,

73,59,27,51,86,87,13,69,57,61

all the selected numbers are between 00to 80except 86,87.

So, 8 toast lie butter side down and 2 toast lie butter side up.

05

Part (c) Step 1:Given Information

We have been given that,

Maria wants to perform a simulation , the probability that 4or fewer pieces of toast out of 4would land butter side down. if the researchers’ probability value is correct =0.81

probability of toast landing butter side down =0.81

06

Part (c) Step 2:Explanation

Number of toast used for simulation=10

Probability of toast falling butter side down =0.81

Number of toast falling butter side down = 0.81*10=8.1

approximately,8toast fall butter side down.

In the given dotplot of 50simulated trials of dropping 10pieces of toast ,8consists of maximum number of dots.

So,there is no convincing evidence that the researchers’ 0.81probability value is incorrect.

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