What is the k? 5/k = 15/20
1 answer:
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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 = ~ N(0,1)
where, = population mean replacement time = 3.3 years
= 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 3.1 years)
P(M 3.1 years) = P(
) = P(Z
-2.357) = 1 - P(Z
2.357)
= 1 - 0.99078 = <u>0.0092</u> or 0.92%
<em>So, in the z table the P(Z </em><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%.