Is 152 points in 8 games 171 points in 9 games equivalent
1 answer:
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This is a problem of Standard Normal distribution.
We have mean= 12 grams
Standard Deviation = 2.5 grams
First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:
So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808
Thus, the <span>probability that the strawberry weighs less than 8.5 grams is 0.0808</span>
We have mean= 12 grams
Standard Deviation = 2.5 grams
First we convert 8.5 to z score. 8.5 converted to z score for given mean and standard deviation will be:
So, from standard normal table we need to find the probability of z score to be less than -1.4. The probability comes out to be 0.0808
Thus, the <span>probability that the strawberry weighs less than 8.5 grams is 0.0808</span>
The probability is 0.0008.
I used a binomial probability on this one, since there are two outcomes (either pass or fail); the events are independent (the probability of one person passing does not influence the probability of another passing); and there are a fixed number of trials (in this case, 20).
For binomial probabilities, we use the formula:
With our information, we have:
I used a binomial probability on this one, since there are two outcomes (either pass or fail); the events are independent (the probability of one person passing does not influence the probability of another passing); and there are a fixed number of trials (in this case, 20).
For binomial probabilities, we use the formula:
With our information, we have: