Use technology or a z-distribution table to find the indicated area.
The weights of grapefruits in a bin are normally distributed with a mean of 261 grams and a standard deviation of 9.4 grams.
Approximately 20% of the grapefruits weigh less than which amount?

239g
247g
253g
261g

Approximately 20% of the grapefruits weigh less than 253.10 grams or 253g.

Step-by-step explanation:

The normal distribution and its parameters

The weights of grapefruits in a bin follow a normal distribution. In a normal distribution, most of the weights are around the mean and less are far from it. The standard deviation measures this. Roughly speaking, the greater the standard deviation is, the normal distribution looks wider than a distribution with a smaller standard deviation. We can say, then, that the mean and the standard deviations determine the shape of the normal distribution.

The standard normal distribution and z-scores

There is a special normal distribution from which we can obtain the probabilities for all normal distributions. In this distribution, the standard deviation plays an important role. We can "transform" any raw data into a z-score. A z-score tells us the distance of a value from the population mean in standard deviations units. A positive z-score indicates that the value is above the mean, and a negative z-score tells us the value is below it.

With a z-score at hand, we can consult a standard normal table, made with the values obtained from a standard normal distribution. This is the special normal distribution mentioned before. Most of the time, this standard normal table shows cumulative probabilities for the corresponding z-scores, and these values come from the cumulative standard normal distribution.

We can obtain the z-scores for any normally distributed raw scores using the next formula:

[1]

Where

x is the raw score.

is the population mean.

is the population standard deviation.

Finding a probability using the standard normal table

For each z-score, there is a corresponding probability in the standard normal table. Thus, for the probability of 20%, there must be a corresponding z-score. To find this, we can consult the standard normal table previously mentioned. In this way, go to the table, look at the probability of 0.20, and try to see which value of z corresponds to this probability. Z-values are at the left and in the first line of the table, we have decimal places for them (0.00, 0.01, 0.02, etc, in case of positive values for z).

Thus, the corresponding value of the z-score for a probability of 20% or 0.20 is about z = -0.84.

If we have a table for only positive values for z in the standard normal table, we know that, since the normal distributions are symmetrical, the corresponding probability that we have to find is 1 - 0.20 = 0.80. That is, the z-score that corresponds to this probability. The z-score that corresponds to a probability of 0.80 is about z = 0.84. However, the value of the raw score is below the mean, and because the symmetry of the normal distribution, z = -0.84. This can be if we study the next expression:

Thus, for a probability of 0.20, we have

In words, the probability for values below z = -0.84 is the same that the probability for values above z = 0.84. This probability is 0.20, approximately.

Finding the value of x

Using [1], we have

The mean is grams.

The standard deviation is grams.

We need to solve the formula for x, which is the raw score we are looking for.

We already know that the value is below the mean (z = -0.84), P(z<-0.84). Then

grams.

Then, approximately 20% of the grapefruits weigh less than 253.10 grams or 253g.  

We can see this probability in the graph below, represented by the shaded area of the normal distribution with a mean of and standard deviation of .

1/5^2 = 7.56/45

y=3n + 1

step-by-step explanation:

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