The normal distribution An automobile battery manufacturer offers a 31/54 warranty on its batteries. The first number in the warranty code is the free-replacement period; the second number is the prorated-credit period. Under this warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charge to the consumer. If the battery fails after 31 months but within 54 months, the manufacturer provides a prorated credit toward the purchase of a new battery. The manufacturer assumes that x, the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months. Use the following Distributions tool to help you answer the questions that follow. (Hint: When you adjust the parameters of a distribution, you must reposition the vertical line (or lines) for the correct areas to be displayed.)
1. If the manufacturer's assumptions are correct, it would reed to replace _______ of its batteries free.
2. The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect. A standard deviation of ____ results in a 1.07% replacement rate.
3. Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

Answer:

1) if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2) a standard deviation of 6.0843 results in a 1.07% replacement rate

3) using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

Step-by-step explanation:

based on the given data;

x will represent the random variable such that the lifetime of its auto batteries, is normally distributed with a mean of 45 months and a standard deviation of 5.6 months

so

x → N( U = 45, ∝ = 5.6)

Under the warranty, if a battery fails within 31 months of purchase, the manufacturer replaces the battery at no charges to the consumer.

if the battery fails after 31 months but within 54 months, the manufacturer provides a prostrated credit towards the purchase of anew battery

1) If the manufacturer's assumptions are correct,

p(x < 3) = p( [x-u / ∝ ] < [ 31-45 / 5.6] )

= p( z < -2.5 )

using the standard normal table,

value of z = 0.0062 ≈ 0.62%

so if the manufacturer's assumptions are correct, it would reed to replace 0.62% of its batteries free.

2)

The company finds that it is replacing 1.07% of its batteries free of charge. It suspects that its assumption standard deviation of the life of its batteries is incorrect, so a standard deviation of ? results in a 1.07%

so lets say;

p ( x < 31 ) = ( 1.07%) = 0.0107

p ( [x-u / ∝ ] < [ 31-45 / ∝] ) = 0.0107

now from the standard table

-2.301 is 1.07%

so

( 31 - 45 / ∝ ) = -2.301

-14 / ∝ = -2.301

∝ = -14 / - 2.301

∝ = 6.0843

therefore a standard deviation of 6.0843 results in a 1.07% replacement rate

3)

Using the revised standard deviation for battery life, what percentage of the manufacturer's batteries don't free replacement but do qualify for the prorated credit?

p( 31 < x < 54 ) = p ( [31 - u / ∝ ] < [ x-u / ∝]  < [ 54 - 45 / ∝] )

= p ( [31 - 45 / 6.0843 ] < [ x-u / ∝]  < [ 54 - 45 / 6.0843] )

= p ( -2.301 < z < 1.4792 )

= p(Z < 1.5) - p(Z < -2.3)

= 0.9393 - 0.0108

= 0.919 ≈ 91.9%

therefore using the revised standard deviation for battery life, 91.9% of the manufacturer's batteries don't get free replacement but qualifies for the prorated credit

H0: μ1 – μ2 = 0 vs. H1: μ1– μ2 ≠ 0
H0: μ1 – μ2 ≠ 0 vs. H1: μ1– μ2 = 0

multiple choice 1
a
b


(b) Specify the decision rule with respect to the p-value.

Reject the null hypothesis if the p-value is
less than
.10.

(c) Find the test statistic tcalc. (A negative value should be indicated by a minus sign. Round your answer to 3 decimal places.)

tcalc


(d) Assume unequal variances to find the p-value. (Use the quick rule to determine degrees of freedom. Round your answer to 4 decimal places.)

p-value


(e) Make a decision.

We
do not reject
the null hypothesis.

(f) State your conclusion.

We
cannot
conclude that there is a difference between the sizes of homes in the two neighborhoods.

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Graph B

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Gender and Residence of Customers
Males Females
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Dorm 295 276
With Parent(s) 55 68
Sorority/Fraternity House 132 204
Other 180 99
What is the probability that a customer is female and lives in a dorm or is female and lives in 'Other'? Express your answer as a fraction or a decimal number rounded to four decimal places.

Answer two questions about Equations AAA and BBB:

\begin{aligned} A.&&5x&=3x \\\\ B.&&5&=3 \end{aligned}

A.

B.

​


​


5x

5

​


=3x

=3

​


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Choose 1 answer:

Choose 1 answer:


(Choice A)

A

Add/subtract the same quantity to/from both sides


(Choice B)

B

Add/subtract a quantity to/from only one side


(Choice C)

C

Multiply/divide both sides by the same non-zero constant


(Choice D)

D

Multiply/divide both sides by the same variable expression

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Choose 1 answer:

Choose 1 answer:


(Choice A)

A

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(Choice B)

B

No

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