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3 years ago

Jeannie's scores on her first four tests were 80, 65, 87, and 75. What will

1 answer:

Ivan3 years ago

6 0

She will have to score 93.
(80+65+87+75+x)/5 = 80
307+x = 400
x =400-307=93
Another way, easier to do in your head:
Notice that 65 is 15 below 80, 87 is 7 above 80, and 75 is 5 below 80. The overall situation is 13 below 80 (because -15+7-5=-13). For a number to push the average back to 80, it must be 13 higher than 80. 80+13=93

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0.0193 = 1.93% probability that there is equipment damage to the payload of at least one of five independently dropped parachutes.

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For each parachute, there are only two possible outcomes. Either there is damage, or there is not. The probability of there being damage on a parachute is independent of any other parachute, which means that the binomial probability distribution is used to solve this question.

To find the probability of damage on a parachute, the normal distribution is used.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

Probability of a parachute having damage.

The opening altitude actually has a normal distribution with mean value 185 and standard deviation 32 m, which means that $\mu = 185, \sigma = 32$

Equipment damage will occur if the parachute opens at an altitude of less than 100 m, which means that the probability of damage is the p-value of Z when X = 100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{100 - 185}{32}$