Question

A local bakary has determined a probability distribution for the number of cheesecakes that they sell in a given day:

X= #sold 0 5 10 15 20
Probability 0.25 0.47 0.12 ? 0.15

a. What is the probability of selling 15 cheesecakes in a given day?
b. What is the probability of selling at least 10 cheesecakes?
c. What is the probability of selling 5 or 15 chassescakes?
d. What is the peobaility of selling 25 cheesecakes?
e. Give the expected number of cheesecakes sold in a day using the discrete probability distribution?
f. What is the probability of selling at most 10 cheesecakes?

Answer 1

Answer:

Step-by-step explanation:

The pdf of  the number of cheesecakes that they sell in a given day: by a bakery are as follows:

X     0         5       10      15     20

P    0.25    0.47   0.12   k     0.15

Use total prob =1, to find k .  

k = 0.01

a)  the probability of selling 15 cheesecakes in a given day = 0.01

b. What is the probability of selling at least 10 cheesecakes =P(X≥10)

= 0.28

c.  the probability of selling 5 or 15 chassescakes P(5)++P(15)=0.48

d.  the probaility of selling 25 cheesecakes = 0

e. Give the expected number of cheesecakes sold in a day using the discrete probability distribution?

=$\Sigma x_i p_i\\=0+2.35+1.2+0.15+3\\=6.7$

f. the probability of selling at most 10 cheesecakes

=P(X≤10)

= 0.84

Answer 2

From the probability distribution, we have that:

a) 0.01 = 1% probability of selling 15 cheesecakes in a given day.

b) 0.28 = 28% probability of selling at least 10 cheesecakes.

c) 0.48 = 48% probability of selling 5 or 15 cheesecakes.

d) 0% probability of selling 25 cheesecakes.

e) The expected number is of 6.7 cheesecakes.

f) 0.84 = 84% probability of selling at most 10 cheesecakes.

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Item a:

The sum of the probabilities of all outcomes has to be 1, and the probability of selling 15 cheesecakes is x.

Applying the sum:

$0.25 + 0.47 + 0.12 + x + 0.15 = 1$

$0.99 + x = 1$

$x = 0.01$

0.01 = 1% probability of selling 15 cheesecakes in a given day.

Item b:

• At least 10 is 10, 15 or 20, that is:

$P(X \geq 10) = P(X = 10) + P(X = 15) + P(X = 20)$

From the values on the table:

$P(X \geq 10) = 0.12 + 0.01 + 0.15 = 0,28$

0.28 = 28% probability of selling at least 10 cheesecakes.

Item c:

This is:

$p = P(X = 5) + P(X = 15) = 0.47 + 0.01 = 0.48$

0.48 = 48% probability of selling 5 or 15 cheesecakes.

Item d:

25 is not part of the distribution, thus, 0% probability of selling 25 cheesecakes.

Item e:

• The expected number is the sum of the multiplication of each outcome by it's respective probability, thus:

$E(X) = 0(0.25) + 5(0.47) + 10(0.12) + 15(0.01) + 20(0.15) = 6.7$

The expected number is of 6.7 cheesecakes.

Item f:

This is:

$P(X \leq 10) = P(X = 0) + P(X = 5) + P(X = 10)$

Substituting the values:

$P(X \leq 10) = 0.25 + 0.47 + 0.12 = 0.84$

0.84 = 84% probability of selling at most 10 cheesecakes.

A similar problem is given at https://brainly.com/question/24188569