PLEASE HELP i need help

Suppose that a high school marching band has 86 members. Of these 86 band members, 26 are seniors, 17 play the trumpet, and 4 ar

GalinKa [24]

Answer: 0.3023

Step-by-step explanation:

86 band members

26 are senior, 17 plays trumpet. If 4 are seniors and play trumpet, we solve this using the bayes theorem of conditional probability.

Probability of choosing band member that plays trumpet =17/86 = 2/43.

Probability of choosing a senior that plays trumpet = 4/26

Probability that a chosen member is a senior given that he plays trumpet = (2/43)/(4/26)

= 0.3023.

8 0

3 years ago

A teacher wrote the equation 3y + 12 = 6x on the board. For what value of b would the additional equation 2y = 4x + b form a sys

guajiro [1.7K]

3y + 12 = 6x
3y = 6x - 12
y = 2x - 4

2y = 4x + (-8)
y = 2x + (-4)....the same as y = 2x - 4

b would have to be a -8

3 0

2 years ago

Read 2 more answers

What is the factored form of 27 - 530?

fenix001 [56]

Answer:

1,2,3

Step-by-step explanation:

8 0

3 years ago

To be eligible for a basketball tournament, a basketball team must win at least 60% of its remaining games. If the team has 21 g

Slav-nsk [51]

IT IS 13 BECAUSE 60 PERCENT OF 20 IS 13

8 0

3 years ago

Let z denote a random variable that has a standard normal distribution. Determine each of the probabilities below. (Round all an

Gelneren [198K]

Answer:

(a) P (Z < 2.36) = 0.9909 (b) P (Z > 2.36) = 0.0091

(c) P (Z < -1.22) = 0.1112 (d) P (1.13 < Z > 3.35) = 0.1288

(e) P (-0.77< Z > -0.55) = 0.0705 (f) P (Z > 3) = 0.0014

(g) P (Z > -3.28) = 0.9995 (h) P (Z < 4.98) = 0.9999.

Step-by-step explanation:

Let us consider a random variable, $X \sim N (\mu, \sigma^{2})$ , then $Z=\frac{X-\mu}{\sigma}$ , is a standard normal variate with mean, E (Z) = 0 and Var (Z) = 1. That is, $Z \sim N (0, 1)$ .

In statistics, a standardized score is the number of standard deviations an observation or data point is above the mean. The z-scores are standardized scores.

The distribution of these z-scores is known as the standard normal distribution.

(a)

Compute the value of P (Z < 2.36) as follows:

P (Z < 2.36) = 0.99086

≈ 0.9909

Thus, the value of P (Z < 2.36) is 0.9909.

(b)

Compute the value of P (Z > 2.36) as follows:

P (Z > 2.36) = 1 - P (Z < 2.36)

= 1 - 0.99086

= 0.00914

≈ 0.0091

Thus, the value of P (Z > 2.36) is 0.0091.

(c)

Compute the value of P (Z < -1.22) as follows:

P (Z < -1.22) = 0.11123

≈ 0.1112

Thus, the value of P (Z < -1.22) is 0.1112.

(d)

Compute the value of P (1.13 < Z > 3.35) as follows:

P (1.13 < Z > 3.35) = P (Z < 3.35) - P (Z < 1.13)

= 0.99960 - 0.87076

= 0.12884

≈ 0.1288

Thus, the value of P (1.13 < Z > 3.35) is 0.1288.

(e)

Compute the value of P (-0.77< Z > -0.55) as follows:

P (-0.77< Z > -0.55) = P (Z < -0.55) - P (Z < -0.77)

= 0.29116 - 0.22065

= 0.07051

≈ 0.0705

Thus, the value of P (-0.77< Z > -0.55) is 0.0705.

(f)

Compute the value of P (Z > 3) as follows:

P (Z > 3) = 1 - P (Z < 3)

= 1 - 0.99865

= 0.00135

≈ 0.0014

Thus, the value of P (Z > 3) is 0.0014.

(g)

Compute the value of P (Z > -3.28) as follows:

P (Z > -3.28) = P (Z < 3.28)

= 0.99948

≈ 0.9995

Thus, the value of P (Z > -3.28) is 0.9995.

(h)

Compute the value of P (Z < 4.98) as follows:

P (Z < 4.98) = 0.99999

≈ 0.9999

Thus, the value of P (Z < 4.98) is 0.9999.

**Use the z-table for the probabilities.

3 0

2 years ago