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The amount of time Ricardo spends brushing his teeth follows a Normal distribution with unknown mean and standard deviation. Ricardo spends less than 1 minute brushing his teeth about 40% of the time. He spends more than 2 minutes brushing his teeth 2% of the time.

Determine the mean and standard deviation of this distribution.

A) Mean = 0.891 minutes, standard deviation = 0.435 minutes.
B) Mean = 1.109 minutes, standard deviation = 2.551 minutes.
C) Mean = 1.891 minutes, standard deviation = 0.435 minutes.
D) Mean = 0.861 minutes, standard deviation = 0.556 minutes.
E) Mean = 1.109 minutes, standard deviation = 0.435 minutes.

User Sammuel
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1 Answer

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E) Mean = 1.109 minutes, standard deviation = 0.435 minutes

Step-by-step explanation

1. Let's denote the mean as μ and the standard deviation as σ.

2. From the given information, we know that Ricardo spends less than 1 minute brushing his teeth about 40% of the time. This implies that the cumulative probability up to 1 minute is 0.4: P(X ≤ 1) = 0.4.

3. We also know that Ricardo spends more than 2 minutes brushing his teeth 2% of the time. This implies that the cumulative probability beyond 2 minutes is 0.02: P(X > 2) = 0.02.

4. Using a standard Normal distribution table or calculator, we can find the corresponding z-scores for these probabilities:

- For P(X ≤ 1) = 0.4, the z-score is approximately -0.253.

- For P(X > 2) = 0.02, the z-score is approximately 2.055.

5. Now, we can use the z-score formula to find the mean and standard deviation:

- For P(X ≤ 1) = 0.4, the z-score formula is: z = (1 - μ) / σ = -0.253.

- For P(X > 2) = 0.02, the z-score formula is: z = (2 - μ) / σ = 2.055.

6. We can solve these two equations simultaneously to find the values of μ and σ.

7. Solving the equations, we find that the mean is approximately 1.109 minutes and the standard deviation is approximately 0.435 minutes.

Therefore, the correct answer is:

E) Mean = 1.109 minutes, standard deviation = 0.435 minutes.

User Nicost
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