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diamong [38]
3 years ago
5

Sarah was trying to decide what kind of cat food to buy at the store. smiley cat costs 7.90 for 10 pounds. purr plus cost .85 pe

r pound. which brand is the better buy?
Mathematics
1 answer:
Keith_Richards [23]3 years ago
5 0
Smiley's cat that is 7.90 for 10 pounds. Because purr plus in total of 10 pounds is 8.50
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The manager of a computer retails store is concerned that his suppliers have been giving him laptop computers with lower than av
34kurt

Answer:

Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is 0.0092.

Yes. The probability of this data is unlikely to have occurred by chance alone.

Step-by-step explanation:

We are given that the replacement times for the model laptop of concern are normally distributed with a mean of 3.3 years and a standard deviation of 0.6 years.

He then randomly selects records on 50 laptops sold in the past and finds that the mean replacement time is 3.1 years.

<em>Let M = sample mean replacement time</em>

The z-score probability distribution for sample mean is given by;

            Z = \frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} }  ~ N(0,1)

where, \mu = population mean replacement time = 3.3 years

            \sigma = standard deviation = 0.6 years

            n = sample of laptops = 50

The Z-score measures how many standard deviations the measure is away from the mean. After finding the Z-score, we look at the z-score table and find the p-value (area) 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.

Now, Probability that the 50 randomly selected laptops will have a mean replacement time of 3.1 years or less is given by = P(M \leq 3.1 years)

 P(M \leq 3.1 years) = P( \frac{ M-\mu}{\frac{\sigma}{\sqrt{n} } }} } \leq \frac{ 3.1-3.3}{\frac{0.6}{\sqrt{50} } }} } ) = P(Z \leq -2.357) = 1 - P(Z \leq 2.357)

                                                           = 1 - 0.99078 = <u>0.0092</u>  or  0.92%          

<em>So, in the z table the P(Z </em>\leq<em> x) or P(Z < x) is given. So, the above probability is calculated by looking at the value of x = 2.357 in the z table which will lie between x = 2.35 and x = 2.36 which has an area of 0.99078.</em>

Hence, the required probability is 0.0092 or 0.92%.

Now, based on the result above; <u>Yes, the computer store has been given laptops of lower than average quality</u> because the probability of this data is unlikely to have occurred by chance alone as the probability of happening the given event is very low as 0.92%.

8 0
3 years ago
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