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Nadya [2.5K]
3 years ago
6

A certain company makes 12-volt car batteries. After many years of product testing, the company knows that the average life of a

battery is normally distributed, with a mean of 50 months and a standard deviation of 9 months. If the company does not want to make refunds for more than 10% of its batteries under the full-refund guarantee policy, for how long should the company guarantee the batteries (to the nearest month)
Mathematics
2 answers:
eduard3 years ago
7 0

Answer:

The company should guarantee the batteries (to the nearest month) for <em>38 months</em>.

Step-by-step explanation:

We have here a <em>normally distributed data</em>. The random variable is the <em>average life of the batteries</em>.

From question, we can say that this random variable has a <em>population mean of 50 months</em> and <em>population standard deviation of 9 months</em>. We can express this mathematically as follows:

\\ \mu = 50 months.

\\ \sigma = 9 months.

The distribution of the random variable (<em>the average life of the batteries</em>) is the normal distribution, and it is determined by two parameters, namely, the mean \\ \mu and \\ \sigma, as we already know.

For the statement: "The company does not want to make refunds for more than 10% of its batteries under the full-refund guarantee policy", we can say that it means that we have determine, first, <em>how many months last less of 10% of the batteries that its average life follows a normal distribution or are normally distributed?</em>

To find this probability, we can use <em>the standard normal distribution</em>, which has some advantages: one of the most important is that we can obtain the probability of any normally distributed data using standardized values given by a z-score, since this distribution (the normal standard) has a mean that equals 0 and standard distribution of 1.

Well, the z-score is given by the formula:

\\ z = \frac{x - \mu}{\sigma} [1]

Where, <em>x</em> is a raw score coming from a normally distributed data. This is the value that we have to transform into a z-score, that is, in a standardized value.

However, from the question, we want to know what value of z represents a cumulative probability of 10% in the <em>cumulative standard normal distribution</em>. We can find it using the <em>standard normal table</em>, available in Statistics books or on the Internet (of course, we can use also Statistics packages or even spreadsheets to find it).

Then, the value of z is, approximately, -1.28, using a cumulative standard normal table for negative values for z. If the cumulative standard normal only has positive values for z, we can obtain it, using the following:

\\ P(z

That is, P(z<-1.28) = P(z>1.28). The probability for P(z<1.28) is approximately, 90%.

Therefore, using the formula [1]:

\\ z = \frac{x - \mu}{\sigma}

\\ -1.28 = \frac{x - 50}{9}

 \\ -1.28 * 9 = x - 50

 \\ -11.52 = x - 50

\\ -11.52 + 50 = x

\\ 38.48 = x

\\ x = 38.48 months.

That is, less than 10% of the batteries have a average life of 38.48 months. Thus, the company should guarantee the batteries (to the nearest month) for 38 months.

Brrunno [24]3 years ago
4 0

Answer:

A certain company makes 12-volt car batteries. After many years of product testing, the company knows that the average life of a battery is normally distributed, with a mean of 50 months and a standard deviation of 9 months. If the company does not want to make refunds for more than 10% of its batteries under the full-refund guarantee policy, for how long should the company guarantee the batteries?

The company should guarantee the batteries for 38 months.

Step-by-step explanation:

Using standard normal table,

P(Z < z) = 10%

=(Z < z) = 0.10

= P(Z <- 1.28 ) = 0.10

z = -1.28

Using z-score formula  

x = zσ  + μ

x = -1.28 *9+50

x = 38

Therefore, the company should guarantee the batteries for 38 months.

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