Correct Ans:
Option A. 0.0100

Solution:
We are to find the probability that the class average for 10 selected classes is greater than 90. This involves the utilization of standard normal distribution.

First step will be to convert the given score into z score for given mean, standard deviation and sample size and then use that z score to find the said probability. So converting the value to z score:

$z-score= \frac{90-83.6}{ \frac{8.7}{ \sqrt{10} } } \\ \\ 
z-score =2.326$

So, 90 converted to z score for given data is 2.326. Now using the z-table we are to find the probability of z score to be greater than 2.326. The probability comes out to be 0.01.

Therefore, there is a 0.01 probability of the class average to be greater than 90 for the 10 classes.

5 0

3 years ago

A roulette wheel has the numbers 1 through 36, 0, and 00. A bet on three numbers pays 11 to 1 (that is, if you bet $1 and one of

OLga [1]

Answer:

You are expected to lose $0.05 (or win -$0.05)

Step-by-step explanation:

Since the roulette wheel has the numbers 1 through 36, 0, and 00, there are 38 possible outcomes.

In this bet, you are allowed to pick 3 out of the 38 numbers. Thus, your chances of winning (P(W)) and losing (P(L)) are:

$P(W)=\frac{3}{38}\\P(L) = 1 - P(W)\\P(L) = \frac{35}{38}\\$

The expected value of the bet is given by the sum of the product of each outcome pay by its probability. Winning the bet means winning $11 while losing the bet means losing $1. The expected value is:

$EV = (11*\frac{3}{38}) -(1*\frac{35}{38})\\EV = -\$0.0525$

Therefore, with a $1 bet, you are expected to lose roughly $0.05

5 0

2 years ago