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brilliants [131]
3 years ago
5

Optimal-Eats blender has a mean time before failure of 40 months with a standard deviation of 6 months, and the failure times ar

e normally distributed. What should be the warranty period, in months, so that the manufacturer will not have more than 10% of the blenders returned
Mathematics
1 answer:
GarryVolchara [31]3 years ago
7 0

Answer:

The warranty period should be of 32 months.

Step-by-step explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Mean time before failure of 40 months with a standard deviation of 6 months.

This means that \mu = 40, \sigma = 6

What should be the warranty period, in months, so that the manufacturer will not have more than 10% of the blenders returned?

The warranty period should be the 10th percentile, which is X when Z has a p-value of 0.1, so X when Z = -1.28.

Z = \frac{X - \mu}{\sigma}

-1.28 = \frac{X - 40}{6}

X - 40 = -1.28*6

X = 32

The warranty period should be of 32 months.

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Use the explicit formula an = a1 + (n – 1) • d to find the 250th term of the sequence below.
solmaris [256]
For this case we have the following equation:
 an = a1 + (n - 1) • d
 Where,
 a1 = 57
 d = 66-57 = 9
 For the term number 250 we have:
 n = 250
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 a250 = 57 + (250 - 1) * 9
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D.2298
8 0
3 years ago
A survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10. (a)
skad [1K]

Answer:

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b) The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

<u>Step-by-step explanation</u>:

Step:-(i)

Given data a survey of 25 young professionals fond that they spend an average of $28 when dining out, with a standard deviation of $10

The sample size 'n' = 25

The mean of the sample   x⁻  = $28

The standard deviation of the sample (S) = $10.

Level of significance ∝=0.05

The degrees of freedom γ =n-1 =25-1=24

tabulated value t₀.₀₅ = 2.064

<u>Step 2:-</u>

The 95% of confidence intervals for the average spending

((x^{-} - t_{\alpha } \frac{S}{\sqrt{n} } ,x^{-} + t_{\alpha }\frac{S}{\sqrt{n} } )

(28 - 2.064 \frac{10}{\sqrt{25} } ,28 + 2.064\frac{10}{\sqrt{25} } )

( 28 - 4.128 , 28 + 4.128)

(23.872 , 32.128)

a) The 95% of confidence intervals for the average spending

(23.872 , 32.128)

b)

Null hypothesis: H₀:μ<30

Alternative Hypothesis: H₁: μ>30

level of significance ∝ = 0.05

The test statistic

t = \frac{x^{-}-mean }{\frac{S}{\sqrt{n} } }

t = \frac{28-30 }{\frac{10}{\sqrt{25} } }

t = |-1|

The calculated value t= 1< 1.711( single tailed test) at 0.05 level of significance with 24 degrees of freedom.

The null hypothesis is accepted

<u>Conclusion</u>:-

The null hypothesis is accepted

A survey of 25 young professionals statistically that the population mean is less than $30

8 0
3 years ago
(( I will give 25 points and brainy for right answers (Please answer all!!) CORRECTLY~!!!))
Alborosie
<span>1. D. Concave Hexagon

2.
(n - 2)180 = 2160
n - 2 = 12
n = 14
</span>
<span>C. 14

3.
</span><span><span>(n - 2)180 = 3060
n - 2 = 17
n = 19
</span>
<span>D. 19

4.
[(n - 2)180]/n =

</span></span><span>= [(15 - 2)180]/15

= 13(180)/15

= 156
</span>
<span>C. 156°</span>
8 0
3 years ago
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