Question

You work for a consumer watchdog publication and are testing the advertising claims of a tire manufacturer. The manufacturer claims that the life spans of the tires are normally distributed, with a mean of 40,000 miles and a standard deviation of 40004000 miles. You test 16 tires and get the following life spans. Complete parts (a) through (c) below. 48,819 40,986 28,175 37,525 29,515 41,602 38,865 37,866 23,405 30,245 38,601 33,815 30,521 37,289 37,018 44,084

(a) Draw a frequency histogram to display these data. Use five classes. Choose the correct answer below.

Is it reasonable to assume that the life spans are normally distributed? Why? Choose the correct answer below.

A. Yes, because the histogram is symmetric and bell-shaped.

B. Yes, because the histogram is neither symmetric nor bell-shaped.

C. No, because the histogram is symmetric and bell-shaped.

D. No, because the histogram is neither symmetric nor bell-shaped.

(b) Find the mean of your sample. (Round to one decimal place as needed.)

Find the standard deviation of your sample. (Round to one decimal place as needed.)

Answer #1

23,405 |

28,175 |

29,515 |

30,245 |

30,521 |

33,815 |

37,018 |

37,289 |

37,525 |

37,866 |

38,601 |

38,865 |

40,986 |

41,602 |

44,084 |

48,819 |

Range=48819-23405=25414

No. of class=5

Class width=25414/5=5082.8

so taking Class width=5100

Frequency Distribution

Class Width | Frequency |

23400-28500 | 2 |

28500-33600 | 3 |

33600-38700 | 6 |

38700-43800 | 3 |

43800-48900 | 2 |

a.) Frequency Histogram

Ans: A.Yes, because the histogram is symmetric and bell-shaped.

b.)Mean=36146

Standard Deviation=5626.3 (using excel.)

A manufacturer claims that the life span of its tires is
52,000 miles. You work for a consumer protection
agency and you are testing these tires. Assume the life spans of
the tires are normally distributed. You select 100
tires at random and test them. The mean life span is
51.831 miles. Assume sigma = 800.
Complete parts (a) through (c).
(a) Assuming the manufacturer's claim is correct, what
is the probability that the mean of the sample is
51,831...

A manufacturer claims that the life span of its tires is
51,000
miles. You work for a consumer protection agency and you are
testing these tires. Assume the life spans of the tires are
normally distributed. You select
100
tires at random and test them. The mean life span is
50,723
miles. Assume
sigmaσequals=800.
Complete parts (a) through (c).
(a) Assuming the manufacturer's claim is correct, what is the
probability that the mean of the sample is
50 comma 72350,723...

A manufacturer claims that the life span of its tires is
48,000
miles. You work for a consumer protection agency and you are
testing these tires. Assume the life spans of the tires are
normally distributed. You select
one hundred
tires at random and test them. The mean life span is
47,858
miles. Assume
sigmaσequals=900
Complete parts (a) through (c).
(a) Assuming the manufacturer's claim is correct, what is the
probability that the mean of the sample is
47,858
miles...

A manufacturer claims that the life span of its tires is
52 comma 00052,000
miles. You work for a consumer protection agency and you are
testing these tires. Assume the life spans of the tires are
normally distributed. You select
100100
tires at random and test them. The mean life span is
51 comma 79951,799
miles. Assume
sigma?equals=700700.
Complete parts? (a) through? (c).
?(a) Assuming the? manufacturer's claim is? correct, what is the
probability that the mean of the sample...

A manufacturer claims that the life span of its tires is 52
comma 000 miles. You work for a consumer protection agency and you
are testing these tires. Assume the life spans of the tires are
normally distributed. You select 100 tires at random and test them.
The mean life span is 51 comma 729 miles. Assume sigmaequals900.
Complete parts (a) through (c). (a) Assuming the manufacturer's
claim is correct, what is the probability that the mean of the
sample...

A tire manufacturer claims that his tires have a mean life of
60,000 miles when used under normal driving conditions. A firm that
requires a larger number of these tires wants to test the claim. If
the claim is correct, the firm will purchase the manufacturer’s
tires; otherwise, the firm will seek another supplier. Now a random
sample of 100 tires is taken and the mean and standard deviation of
the 100 tires are found. Using these sample results, a...

A tire manufacturer claims that his tires have a mean life of
60,000 miles when used under normal driving conditions. A firm that
requires a larger number of these tires wants to test the claim. If
the claim is correct, the firm will purchase the manufacturer’s
tires; otherwise, the firm will seek another supplier. Now a random
sample of 100 tires is taken and the mean and standard deviation of
the 100 tires are found. Using these sample results, a...

A tire manufacturer claims that his tires have a mean life of
60,000 miles when used under normal driving conditions. A firm that
requires a larger number of these tires wants to test the claim. If
the claim is correct, the firm will purchase the manufacturer’s
tires; otherwise, the firm will seek another supplier. Now a random
sample of 100 tires is taken and the mean and standard deviation of
the 100 tires are found. Using these sample results, a...

A tire manufacturer has been producing tires with an average
life expectancy of 26,000 miles. Now the company is advertising
that its new tires' life expectancy has increased.
In order to test the legitimacy of the advertising campaign, an
independent testing agency tested a sample of 8 of their tires and
has provided the following data.
Life Expectancy
(In Thousands of Miles)
28
27
25
26
28
26
29
25
a.
Determine the mean and the standard deviation.
b....

ADVERTISEMENT

Get Answers For Free

Most questions answered within 1 hours.

ADVERTISEMENT

asked 2 minutes ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 1 hour ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago

asked 2 hours ago