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artcher [175]
3 years ago
10

The table describes a probability distribution for a certain event. Find the mean, 4, for the probability distribution.

Mathematics
1 answer:
Vikentia [17]3 years ago
3 0

Answer:

C. 19.2 is the mean for this probability distribution.

Step-by-step explanation:

The mean for a probability distribution is ∑ P(x).x

Here, x∈{10, 20, 30, 40, 50}

P(10) × 10= 5.34

P(20) × 20=3.86

P(30) × 30=4.2

P(40) × 40=3.4

P(50) × 50=2.4

∑=5.34+3.86+4.2+3.4+2.4= 19.2

Therefore, the mean of this probability distribution is 19.2.

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A particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams. The heavies
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Answer:

The heaviest 5% of fruits weigh more than 747.81 grams.

Step-by-step explanation:

We are given that a particular fruit's weights are normally distributed, with a mean of 733 grams and a standard deviation of 9 grams.

Let X = <u><em>weights of the fruits</em></u>

The z-score probability distribution for the normal distribution is given by;

                              Z  =  \frac{X-\mu}{\sigma}  ~ N(0,1)

where, \mu = population mean weight = 733 grams

            \sigma = standard deviation = 9 grams

Now, we have to find that heaviest 5% of fruits weigh more than how many grams, that means;

                    P(X > x) = 0.05       {where x is the required weight}

                    P( \frac{X-\mu}{\sigma} > \frac{x-733}{9} ) = 0.05

                    P(Z > \frac{x-733}{9} ) = 0.05

In the z table the critical value of z that represents the top 5% of the area is given as 1.645, that means;

                               \frac{x-733}{9}=1.645

                              {x-733}}=1.645\times 9

                              x = 733 + 14.81

                              x = 747.81 grams

Hence, the heaviest 5% of fruits weigh more than 747.81 grams.

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