The weights of a certain brand of candies are normally distributed with a mean weight of 0.8599 g and a standard deviation of 0.0519 g. A sample of these candies came from a package containing 453 candies, and the package label stated that the net weight is 387.0 g. (If every package has 453 candies, the mean weight of the candies must exceed 387.0/453=0.8542 g for the net contents to weigh at least 387.0g.)

a. If 1 candy is randomly selected, find the probability that it weighs more than0.8542g. The probability is
(Round to four decimal places as needed.)

The probability is 0.5438

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean and standard deviation , the zscore of a measure X is given by:

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 pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

a. If 1 candy is randomly selected, find the probability that it weighs more than 0.8542g.

This is 1 subtracted by the pvalue of Z when X = 0.8542. So

has a pvalue of 0.4562

1 - 0.4562 = 0.5438

The probability is 0.5438

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}

step-by-step explanation:

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