2 3/4 ÷ 3 2/3 ⋅ 3 3/4 ÷ 2 1/2

c) Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

Step-by-step explanation:

We use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

A result is considered significantly high if it is more than 2.5 standard deviations above the mean.

Normal probability distribution

Problems of normally distributed distributions can be solved using 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.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$ , $\sigma = \sqrt{V(X)}$ .

Assume that 24.5% of people have sleepwalked.

This means that $p = 0.245$

Sample of 1478 adults:

This means that $n = 1478$

Mean and standard deviation:

$\mu = 1478*0.245 = 362.11$

$\sigma = \sqrt{1478*0.245*0.755} = 16.5346$

a. Assuming that the rate of 24.5% is correct, find the probability that 369 or more of the 1478 adults have sleepwalked.

Using continuity correction, this is $P(X \geq 369 - 0.5) = P(X \geq 368.5)$ , which is 1 subtracted by the p-value of Z when X = 368.5.

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

$Z = \frac{368.5 - 362.11}{16.5346}$

$Z = 0.39$

$Z = 0.39$ has a p-value of 0.6517

1 - 0.6517 = 0.3483

0.3483 = 34.83% probability that 369 or more of the 1478 adults have sleepwalked.

b. Is that result of 369 or more significantly high?

362.11 + 2.5*16.5346 = 403.4

369 < 403.4, which means that 369 is less than 2.5 standard deviations above the mean, and thus, a result of 369 is not significantly high.

c. What does the result suggest about the rate of 24.5%?

Since the sample result is not significant, it suggests that the rate of 24.5% is a good estimate for the percentage of people that have sleepwalked.

3 0

3 years ago

A merchant bought 300 apples at the rate of 5 for RS.90 and sold 120 of them at Rs.300 per dozen. If he wants to make a profit o

dangina [55]

Answer:

I think he should continue with the same rate of Rs300 per dozen to gain profit of Rs2100

At the rate of Rs 90 pre 5 apples he have to pay Rs 5400 for 300(25 dozen) apples.

total he have 25 dozen of apples.if he sold 12 dozen of appels at 300 per dozen then he earns Rs 3600. according to the merchant goal he wanted to make profit of 2100.we will add rs 2100 to CP hence we will get SP of Rs 5400+2100=7500. if he had already sell 12 dozens at rs 3600 then we will reduce the amount from SP hence we will get Rs 3900 to be earn. 13 dozens of apple is left so we can divide 3900/13 and we will get Rs 300.

if he continues selling those remaining 13 dozens of apple at the same rate the he can make profit of rs 2100.

6 0

2 years ago

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Kazeer [188]

Well you have to draw it but I need numbers like 50 or something like that how much was he paid?

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3 years ago