Please help me answer this Statistics question.?

Are attitudes toward shopping changing? Sample surveys show that fewer people enjoy shopping than in the past. A recent survey asked a nationwide random sample of 2490 adults if they agreed or disagreed that "I like buying new clothes, but shopping is often frustrating and time-consuming." The population that the poll wants to draw conclusions about is all U.S. residents aged 18 and over. Suppose that in fact 63% of all adult U.S. residents would say "Agree" if asked the same questions. What is the probability that 1558 or more of the sample agree?

Please help me answer this Statistics question.?
Let Xb be the shoppers who say "agree". Xb has the binomial distribution with n = 2490 trials and success probability p = 0.63





In general, if X has the binomial distribution with n trials and a success probability of p then


P[Xb = x] = n!/(x!(n-x)!) * p^x * (1-p)^(n-x)


for values of x = 0, 1, 2, ..., n


P[Xb = x] = 0 for any other value of x.





To use the normal approximation to the binomial you must first validate that you have more than 10 expected successes and 10 expected failures. In other words, you need to have n * p %26gt; 10 and n * (1-p) %26gt; 10.





Some authors will say you only need 5 expected successes and 5 expected failures to use this approximation. If you are working towards the center of the distribution then this condition should be sufficient. However, the approximations in the tails of the distribution will be weaker especially if the success probability is low or high. Using 10 expected successes and 10 expected failures is a more conservative approach but will allow for better approximations especially when p is small or p is large.





In this case you have:


n * p = 2490 * 0.63 = 1568.7 expected success


n * (1 - p) = 2490 * 0.37 = 921.3 expected failures





We have checked and confirmed that there are enough expected successes and expected failures. Now we can move on to the rest of the work.





If Xb ~ Binomial(n, p) then we can approximate probabilities using the normal distribution where Xn is normal with mean μ = n * p, variance σ² = n * p * (1-p), and standard deviation σ





Xb ~ Binomial(n = 2490 , p = 0.63 )


Xn ~ Normal( μ = 1568.7 , σ² = 580.419 )


Xn ~ Normal( μ = 1568.7 , σ = 24.09189 )





I have noted two different notations for the Normal distribution, one using the variance and one using the standard deviation. In most textbooks and in most of the literature, the parameters used to denote the Normal distribution are the mean and the variance. In most software programs, the standard notation is to use the mean and the standard deviation.





The probabilities are approximated using a continuity correction. We need to use a continuity correction because we are estimating discrete probabilities with a continuous distribution. The best way to make sure you use the correct continuity correction is to draw out a small histogram of the binomial distribution and shade in the values you need. The continuity correction accounts for the area of the boxes that would be missing or would be extra under the normal curve.





P( Xb %26lt; x) ≈ P( Xn %26lt; (x - 0.5) )


P( Xb %26gt; x) ≈ P( Xn %26gt; (x + 0.5) )


P( Xb ≤ x) ≈ P( Xn ≤ (x + 0.5) )


P( Xb ≥ x) ≈ P( Xn ≥ (x - 0.5) )


P( Xb = x) ≈ P( (x - 0.5) %26lt; Xn %26lt; (x + 0.5) )


P( a ≤ Xb ≤ b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b + 0.5) )


P( a ≤ Xb %26lt; b ) ≈ P( (a - 0.5) %26lt; Xn %26lt; (b - 0.5) )


P( a %26lt; Xb ≤ b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b + 0.5) )


P( a %26lt; Xb %26lt; b ) ≈ P( (a + 0.5) %26lt; Xn %26lt; (b - 0.5) )





In the work that follows Xb has the binomial distribution, Xn has the normal distribution and Z has the standard normal distribution.





Remember that for any normal random variable Xn, you can transform it into standard units via: Z = (Xn - μ ) / σ











P( Xb ≥ 1558 ) =





2490


∑ P(Xb = x) = 0.679494


x = 1558





≈ P( Xn ≥ 1557.5 )


= P( Z ≥ ( 1557.5 - 1568.7 ) / 24.09189 )


= P( Z ≥ -0.4648868 )


= 0.6789937

shark teeth