In a recent poll, 45% of survey respondents said that, if they only had one child, they would prefer the child to be a boy. Suppose you conducted a survey of 160 randomly selected students on your campus and find that 79 of them would prefer a boy. Complete parts (a) and (b) below. Click here to view the standard normal distribution table (page 1). LOADING... Click here to view the standard normal distribution table (page 2). LOADING... (a) Use the normal approximation to the binomial to approximate the probability that, in a random sample of 160 students, at least 79 would prefer a boy, assuming the true percentage is 45%. The probability that at least 79 students would prefer a boy is 3636. (Round to four decimal places as needed.)

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In a recent​ poll, 45​% of survey respondents said​ that, if they only had one​ child, they would prefer the child to be a boy. Suppose you conducted a survey of 160 randomly selected students on your campus and find that 79 of them would prefer a boy. Complete parts​ (a) and​ (b) below.    Click here to view the standard normal distribution table (page 1).   LOADING...   Click here to view the standard normal distribution table (page 2).  LOADING...  ​ (a) Use the normal approximation to the binomial to approximate the probability​ that, in a random sample of 160 ​students, at least 79 would prefer a​ boy, assuming the true percentage is 45​%.   The probability that at least 79 students would prefer a boy is 3636.  ​ (Round to four decimal places as​ needed.)

Hattie Schaeffer · Accepted Answer

Step 1  Given,  n=160  x=79  p=0.45   The normal approximation of the binomial distribution can be used.  If np(1−p)≥10 is satisfied:  np(1−p)=160⋅0.45⋅(1−0.45)=39.6&amp;gt;10  Above rule is satisfied, thus we can use the normal approximation.   To find,  a. the probability​ that, in a random sample of 160 ​students, at least 79 would prefer a​ boy  Step 2  Answer:  a. The Z - score is the value (using the continuity correction) decreased by the mean np and divided by the standard deviation npq=np(1−p).  Z=x−npnp(1−p)=78.5−(160×0.45)160×0.45×0.55=1.03  Determine the corresponding normal probability using the normal probability table is:  P(X≥79)  =P(X&amp;gt;78.5)  =P(Z&amp;gt;1.03)  =1−P(Z&amp;lt;1.03)  =1−0.8485  =0.1515  The probability that at least 79 students would prefer a boy is 0.1515 .

Dawn Neal · Accepted Answer

Step 1  We have to find p[x≥79]  Rewriting the problem using the continuity correction factor, that is  p[x≥79]=p[x≥79−0.5]=p[x≥78.5]  ⇒p[x≥78.5]=p[x−160×0.45160×0.45×0.55≥78.5−160×0.45160×0.45×0.55]  =p[z≥1.03]  =0.5−p[0≥z≥1.03]  =0.5−0.3485=0.1515=15.15%

In a recent poll, 45% of survey respondents said that, if

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