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slega [8]
2 years ago
10

Mattel Corporation produces a remote-controlled car that requires three AA batteries. The mean life of these batteries in this p

roduct is 34 hours. The distribution of the battery lives closely follows the normal probability distribution with a standard deviation of 5.5 hours. As a part of their testing program Sony tests samples of 25 batteries.
What can you say about the shape of the distribution of sample mean?


What is the standard error of the distribution of the sample mean? (Round your answer to 4 decimal place.)


What proportion of the samples will have a mean useful life of more than 36 hours? (Round your answer to 4 decimal place.)


What proportion of the sample will have a mean useful life greater than 33.5 hours? (Round your answer to 4 decimal place.)


What proportion of the sample will have a mean useful life between 33.5 and 36 hours? (Round your answer to 4 decimal place.)
Mathematics
1 answer:
goldfiish [28.3K]2 years ago
6 0

Answer:

0.0406, 0.8284,0.7887

Step-by-step explanation:

Given that Mattel Corporation produces a remote-controlled car that requires three AA batteries

X is N(34, 5.5)

Hence sample size of 25 would follow a t distribution with df = 24

This is because sample size <30

t distribution with df 24 would be bell shaped symmetrical about the mean and unimodal.

Std error of sample mean = std dev /sqrt n=\frac{5.5}{5} \\=1.1

Prob (X>36) = P(t>\frac{36-34}{1.1} ) = P(t>1.82)\\= 0.04063

i.e nearly 4.1% of the sample would have a mean useful life of more than 36 hours

X>33.5 implies t>-0.45

=0.82837

=0.8284 proportion will have a mean useful life greater than 33.5 hours

Proportion between 33.5 and 36 hours

= 0.3284+0.4593=0.7887

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The correct option is D.) Causation cannot be determined from an observational study.

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Answer:

11

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A local fraternity is conducting a raffle where 55 tickets are to be sold—one per customer. There are three prizes to be awarded
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(a) 0.0152%

(b) 1.1663%

(c) 19.4297%

(d) 79.3787%

Step-by-step explanation:

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(b) The probability that the four organizers win exactly two of the prizes is:

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(c) The probability that the four organizers win exactly one of the prizes is:

P = \frac{4}{55}*\frac{51}{54}*\frac{50}{53}+\frac{51}{55}*\frac{4}{54}*\frac{50}{53}+\frac{51}{55}*\frac{50}{54}*\frac{4}{53}\\P=19.4397\%

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P = \frac{51}{55}*\frac{50}{54}*\frac{49}{53}}\\P=79.3787\%

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