The are six cards in box. 2 of them are blue and 4 are red. You randomly draw one card.
2 answers:
Given Information:
Number of blue cards = 2
Number of red cards = 4
Total number of cards = 6
Required Information:
a) P(Blue) = ?
b) P(Red) = ?
c) which color card has a higher chance of being selected?
Answer:
a) P(Blue) = 1/3
b) P(Red) = 2/3
c) Red card has a higher chance of being selected as compared to Blue card.
Step-by-step explanation:
We know that probability is given by
a)
The probability of selecting a Blue card is,
b)
The probability of selecting a Red card is,
c)
The probability of selecting a Red card is greater (2/3) as compared to Blue card (1/3) therefore, Red card has a higher chance of being selected as compared to Blue card.
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Answer:
a: 28 < µ < 34
Step-by-step explanation:
We need the mean, var, and standard deviation for the data set. See first attached photo for calculations for these...
We get a mean of 222/7 = 31.7143
and a sample standard deviation of: 4.3079
We can now construct our confidence interval. See the second attached photo for the construction steps.
They want a 90% confidence interval. Our sample size is 7, so since n < 30, we will use a t-score. Look up the value under the 10% area in 2 tails column, and degree of freedom is 6 (degree of freedom is always 1 less than sample size for confidence intervals when n < 30)
The t-value is: 1.943
We rounded down to the nearest person in the interval because we don't want to over estimate. It said 28.55, so more than 28 but not quite 29, so if we use 29 as the lower limit, we could over estimate. It's better to use 28 and underestimate a little when considering customer flow.
Answer:
Probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.
Step-by-step explanation:
We are given that a certain car model has a mean gas mileage of 31 miles per gallon (mpg) with a standard deviation 3 mpg.
A pizza delivery company buys 43 of these cars.
<em>Let </em><em> = sample average mileage of the fleet </em>
<em />
The z-score probability distribution of sample average is given by;
Z = ~ N(0,1)
where, = mean gas mileage = 31 miles per gallon (mpg)
= standard deviation = 3 mpg
n = sample of cars = 43
So, probability that the average mileage of the fleet is greater than 30.7 mpg is given by = P(<em> </em>> 30.7 mpg)
P(<em> </em>> 30.7 mpg) = P(
>
) = P(Z > -0.66) = P(Z < 0.66)
= 0.7454
<em>Because in z table area of P(Z > -x) is same as area of P(Z < x). Also, the above probability is calculated using z table by looking at value of x = 0.66 in the z table which have an area of 0.7454. </em>
Therefore, probability that the average mileage of the fleet is greater than 30.7 mpg is 0.7454.