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Ganezh [65]
3 years ago
12

I need help with this question please help me

Mathematics
2 answers:
Mars2501 [29]3 years ago
8 0
By using a Venn diagram
Anastasy [175]3 years ago
7 0
By using a Venn diagram
You might be interested in
A Government company claims that an average light bulb lasts 270 days. A researcher randomly selects 18 bulbs for testing. The s
Fofino [41]

Answer:

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

When the distribution is normal, we use the z-score formula.

In a set with mean \mu and standard deviation \sigma, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean \mu and standard deviation \sigma, the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean \mu and standard deviation s = \frac{\sigma}{\sqrt{n}}.

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

For a proportion p in a sample of size n, the sampling distribution of the sample proportion will be approximately normal with mean \mu = p and standard deviation s = \sqrt{\frac{p(1-p)}{n}}

In this question, we have that:

\mu = 270, \sigma = 90, n = 18, s = \frac{90}{\sqrt{18}} = 21.2

What is the probability that 18 randomly selected bulbs would have an average life of no more than 260 days?

This is the pvalue of Z when X = 260. So

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

By the Central Limit Theorem

Z = \frac{X - \mu}{s}

Z = \frac{260 - 270}{21.2}

Z = -0.47

Z = -0.47 has a pvalue of 0.3192.

31.92% probability that 18 randomly selected bulbs would have an average life of no more than 260 days

5 0
3 years ago
I need help w/ these two problems if you don't mind.
dusya [7]
The answers would be a and then ctell me if you need an explaination

4 0
3 years ago
Read 2 more answers
Using fermat's little theorem, find the least positive residue of $2^{1000000}$ modulo 17.
torisob [31]
Fermat's little theorem states that
a^p≡a mod p

If we divide both sides by a, then
a^{p-1}≡1 mod p
=>
a^{17-1}≡1 mod 17
a^{16}≡1 mod 17

Rewrite
a^{1000000} mod 17  as
=(a^{16})^{62500} mod 17
and apply Fermat's little theorem
=(1)^{62500} mod 17
=>
=(1) mod 17

So we conclude that
a^{1000000}≡1 mod 17

6 0
4 years ago
movie theater sold 5400 tickets yesterday of those, 3240 tickets were for Star Wars. what was the percentage it sold yesterday w
Gnesinka [82]
3240 star wars tickets divided into 5400 total tickets = .6 × 100 = 60%
May the Force be with you.
8 0
3 years ago
Please help its due tonight!!
Lynna [10]

Answer:

your really think ik the answer

Step-by-step explanation:

yea the  sorry

8 0
3 years ago
Read 2 more answers
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