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Licemer1 [7]
4 years ago
6

The mean temperature for the first 7 days in January was 6 degrees.The temperature on the 8 days was 10 degrees . What is the me

an temperature for the first 8 days in January?
Mathematics
1 answer:
kolezko [41]4 years ago
3 0

Answer: The mean temperature for the first eight days is 6.5 degrees

Step-by-step explanation: The most important piece of clue has been given which  is the mean (average) for the observed data set, which is 7 days.

Note that the formula for the mean of a data set is derived as;

Mean = ∑x / f

Where ∑x is the summation of all observed data set and f is the number of data observed, that is 7. The formula now becomes;

6 = ∑x / 7

By cross multiplication, we now have,

6 * 7 = ∑x

42 = ∑x

This means the addition of all temperature observed on the first 7 days is 42. The temperature on the eighth day is now given as 10 degrees, this means the summation of all observed data for the first eight days would become 42 + 10 which equals 52. Therefore when calculating the mean for the first eight days, ∑x is now 52. The formula for the first eight days therefore is derived as follows;

Mean = ∑x / 8

Mean = 52 / 8

Mean = 6.5

The calculations therefore show that the mean temperature for the first eight days in January is 6.5 degrees

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Suppose the weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams. The weights
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Answer:

a) 0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b) The weight that 80% of the apples exceed is of 78.28g.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean \mu and standard deviation \sigma, the z-score of a measure X is given by:

Z = \frac{X - \mu}{\sigma}

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Weights of apples are normally distributed with a mean of 85 grams and a standard deviation of 8 grams.

This means that \mu = 85, \sigma = 8

a. Find the probability a randomly chosen apple exceeds 100 g in weight.

This is 1 subtracted by the p-value of Z when X = 100. So

Z = \frac{X - \mu}{\sigma}

Z = \frac{100 - 85}{8}

Z = 1.875

Z = 1.875 has a p-value of 0.9697

1 - 0.9696 = 0.0304

0.0304 = 3.04% probability a randomly chosen apple exceeds 100 g in weight.

b. What weight do 80% of the apples exceed?

This is the 100 - 80 = 20th percentile, which is X when Z has a p-value of 0.2, so X when Z = -0.84.

Z = \frac{X - \mu}{\sigma}

-0.84 = \frac{X- 85}{8}

X - 85 = -0.84*8

X = 78.28

The weight that 80% of the apples exceed is of 78.28g.

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