A card is drawn at random out of a well-shuffled deck of 52 cards. Find the probability of drawing a six comma seven comma or e
1 answer:
Answer: , or ~23.07%
Step-by-step explanation:
There are 52 cards in your average deck.
For this question, you need the numbers of the cards in the deck. There are four of each number from 2 through 10, and four aces, jacks, queens, and kings.
Considering there are four 6s, four 7s, and four 8s, there are 12 cards that fit the description.
Divide 12 by the total number of cards in the deck, 52, to get your answer.
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Answer:
No, we can't reject the dealer's claim with a significance level of 0.05.
Step-by-step explanation:
We are given that a local car dealer claims that 25% of all cars in San Francisco are blue.
You take a random sample of 600 cars in San Francisco and find that 141 are blue.
<u><em>Let p = proportion of all cars in San Francisco who are blue</em></u>
SO, Null Hypothesis, : p = 25% {means that 25% of all cars in San Francisco are blue}
Alternate Hypothesis, : p
25% {means that % of all cars in San Francisco who are blue is different from 25%}
The test statistics that will be used here is <u>One-sample z proportion</u> <u>statistics</u>;
T.S. = ~ N(0,1)
where, = sample proportion of 600 cars in San Francisco who are blue =
= 0.235
n = sample of cars = 600
So, <u><em>test statistics</em></u> =
= -0.866
<em>Now at 0.05 significance level, the z table gives critical values of -1.96 and 1.96 for two-tailed test. Since our test statistics lies within the range of critical values of z so we have insufficient evidence to reject our null hypothesis as it will not fall in the rejection region due to which we fail to reject our null hypothesis.</em>
Therefore, we conclude that 25% of all cars in San Francisco are blue which means the dealer's claim was correct.