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r-ruslan [8.4K]
3 years ago
11

A quality control specialist for a large computer store conducted a random sample of 150 laptops from its stock. She found 3 did

not turn on, and 5 required system updates to run properly. What is a reasonable probability model for the laptops the store has in stock?
Select all that apply:

2% of the laptops will require system updates.
3.3% of the laptops will require system updates.
92% of the laptops will function properly.
5.3% of the laptops will require system updates.
96.6% of the laptops will function properly.
94.7% of the laptops will function properly.
Mathematics
1 answer:
ANEK [815]3 years ago
8 0

Answer:

<u>3.3% of the laptops will require system updates.  This option applies because 3.3% of 150 is 5.</u>

<u>94.7% of the laptops will function properly. This option applies because 94.7% of 150 is 142.</u>

Step-by-step explanation:

1. Let's review the information given to us to answer the question correctly:

Random sample = 150 laptops

Number of laptops that didn't turn on = 3

Number of laptops that required system updates = 5

Therefore, number of laptops that will function properly = 142

2. What is a reasonable probability model for the laptops the store has in stock?

2% of the laptops will require system updates.  This doesn't apply because 2% of 150 is 3 and not 5.

<u>3.3% of the laptops will require system updates.  This option applies because 3.3% of 150 is 5.</u>

92% of the laptops will function properly. This doesn't apply because 92% of 150 is 138 and not 142.

5.3% of the laptops will require system updates.  We are not totally sure if the laptops that don't turn on, will also require system updates. Most likely, they will but we're not 100% sure. Therefore, this option doesn't apply.

96.6% of the laptops will function properly.  This doesn't apply because 96.6% of 150 is 145 and not 142.

<u>94.7% of the laptops will function properly. This option applies because 94.7% of 150 is 142.</u>

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Step-by-step explanation:

Let's define the following event :

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P(X=6)=\frac{(5)(4)(3)(2)(1)(5)(4!)}{(10!)}=\frac{1}{252}

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