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Measurement error that is normally distributed with a mean of 0 and a standard deviation of 0.5 gram is added to the true weight of a sample. Then the measurement is rounded to the nearest gram. Suppose that the true weight of a sample is 166.0 grams.

(a) What is the probability that the rounded result is 167 grams?
(b) What is the probability that the rounded result is 167 grams or more?

User King Jia
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2 Answers

9 votes
9 votes

Final answer:

The probability that the rounded result is 167 grams is 0. The probability that the rounded result is 167 grams or more is approximately 0.6915, or 69.15%.

Step-by-step explanation:

To calculate the probability that the rounded result is 167 grams, we need to consider the measurement error and rounding. The measurement error is normally distributed with a mean of 0 and a standard deviation of 0.5 gram. When the true weight is 166.0 grams, the measurement error can range from -0.5 gram to 0.5 gram. However, since the result is rounded to the nearest gram, any measurement error in the range of -0.5 to 0.5 gram would be rounded to 0 and the rounded result would be 166 grams. Therefore, the probability that the rounded result is 167 grams is 0.

To calculate the probability that the rounded result is 167 grams or more, we need to consider measurements with an error of 0.5 gram or less. Since the measurement error is normally distributed with a mean of 0 and a standard deviation of 0.5 gram, we can use the cumulative distribution function of the normal distribution to find this probability. The cumulative distribution function (CDF) gives us the probability that a random variable is less than or equal to a given value. In this case, we want to find the probability that the measurement error is less than or equal to 0.5 gram (to make the rounded result 167 grams or more). Plugging the values into the CDF, we get a probability of approximately 0.6915, or 69.15%.

User Liao Zhuodi
by
2.5k points
16 votes
16 votes

Answer:

(a)
0.15731

(b)0.02275

Step-by-step explanation:

We are given that

Mean=0

Standard deviation=0.5 g

True weight of a sample=166 g

Let X denote the normal random variable with mean =166+0=166

(a)

P(166.5<X<167.5)

=
P((166.5-166)/(0.5)<(X-\mu)/(\sigma)<(167.5-166)/(0.5))

=
P(1<Z<3)

=
P(Z<3)-P(Z<1)


=0.99865-0.84134


=0.15731

(b)


P(X>167)=P(Z>(167-166)/(0.5))


=P(Z>2)


=1-P(Z<2)


=1-0.97725


=0.02275

User MichaelCMS
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3.5k points