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IRISSAK [1]
2 years ago
13

Britney just got a new video game for her birthday. Yesterday, she played for 27 minutes and beat 3 levels. Today, her parents s

aid she can play for 45 minutes. If she beats levels at the same rate, how many levels should Britney beat today?
Mathematics
1 answer:
Tom [10]2 years ago
8 0

Answer:

5 levels

Step-by-step explanation:

To know the rate of her beating levels yesterday, divide the # of minutes she played (27 minutes) to the # of levels she beat (3 levels).

  • 27 ÷ 3 = 9

So she played for 9 minutes to beat a single level.

Since she will try to beat levels the same rate as yesterday, you will have to divide the #of minutes her parents let her play (45 minutes) to yesterday's rate (9 minutes per level).

  • 45 ÷ 9 = 5

Britney should beat 5 levels today if she wants to beat levels at the same rate as yesterday's.

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Answer:

(a) E(X) = 950

(b) $ COV = 0.007255$

(c) P(X > 980) = 0.00001\\\\

Step-by-step explanation:

The given problem can be solved using binomial distribution since the product either fails or passes, the probability of failure or success is fixed and there are n repeated trials.

probability of failure = q = 0.05

probability of success = p = 1 - 0.05 = 0.95

number of trials = n = 1000

(a) What is the expected number of non-defective units?

The expected number of non-defective units is given by

E(X) = n \times p \\\\E(X) = 1000 \times 0.95 \\\\E(X) = 950

(b) what is the COV of the number of non-defective units?

The coefficient of variance is given by

$ COV = \frac{\sigma}{E(X)} $

Where the standard deviation is given by

\sigma = \sqrt{n \times p\times q} \\\\\sigma = \sqrt{1000 \times 0.95\times 0.05} \\\\\sigma = 6.892

So the coefficient of variance is

$ COV = \frac{6.892}{950} $

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(c) What is the probability of having more than 980 non-defective units?

We can use the Normal distribution as an approximation to the Binomial distribution since n is quite large and so is p.

P(X > 980) = 1 - P(X < 980)\\\\P(X > 980) = 1 - P(Z < \frac{x - \mu}{\sigma} )\\\\

We need to consider the continuity correction factor whenever we use continuous probability distribution (Normal distribution) to approximate discrete probability distribution (Binomial distribution).

P(X > 980)  = 1 - P(Z < \frac{979.5 - 950}{6.892} )\\\\P(X > 980)  = 1 - P(Z < \frac{29.5}{6.892} )\\\\P(X > 980)  = 1 - P(Z < 4.28)\\\\

The z-score corresponding to 4.28 is 0.99999

P(X > 980) = 1 - 0.99999\\\\P(X > 980) = 0.00001\\\\

So it means that it is very unlikely that there will be more than 980 non-defective units.

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