The Southside Bowling Alley has collected data on the number of children that come to birthday parties held at the bowling alley. Let the random variable \( X = \) the number of children per party. The distribution for the random variable \( X \) is given below.\[\begin{array}{|c|c|c|c|c|c|c|}\hlinek & 5 & 6 & 7 & 8 & 9 & 10 \\\hlineP(X=k) & 0.15 & ? & 0.1 & 0.25 & 0.1 & 0.2 \\\hline\end{array}\]a) The probability that at least 7 children will come to a party is \([ \text{Select} ]\).b) The expected number of children that will attend this party is \( E(X) = [ \text{Select} ] \). \( E(x) = \Sigma k \cdot P(X = k) \). Write answer to two decimal places.c) If the standard deviation for the number of children per party is 1.72, then the variance will be \([ \text{Select} ]\). Write answer to two decimal places.

Given data is k 5 6 7 8 9 10 P(X=k) 0.15 ? 0.1 0.25 0.1 0.2Let "x" be the missing probabilityWe know that sum of the probabilities is equal to one∑P(X=k)=10.15+x+0.1+0.25+0.1+0.2=10.8+x=1x=1-0.8x=0.2 k 5 6 7 8 9 10 P(X=k) 0.15 0.2 0.1 0.25 0.1 0.2 a)The probability that atleast 7 children will come to party is =P(x≥7)=P(x=7)+P(x=8)+P(x=9)+P(x=10)=0.1+0.25+0.1+0.2=0.65The probability that atleast 7 children will come to party is 0.65b)The Expected number of children that will attend this party isE(x)=∑k*P(X=k)=(5×0.15)+(6×0.2)+(7×0.1)+(8×0.25)+(9×0.1)+(10×0.2)=0.75+1.2+0.7+2+0.9+2=7.55E(X)=7.55E(x2)=∑k2*P(X=k)=(52×0.15)+(62×0.2)+(72×0.1)+(82×0.25)+(92×0.1)+(102×0.2)=3.75+7.2+4.9+16+8.1+20=59.95E(X2)=59.95standard deviation=E(x2)-(E(x)2)standard deviation=59.95-(7.552)standard deviation=1.716828471standard deiation≅1.72variance=E(x2)-E(x)2 variance=59.95-(7.55)2 Variance=2.9475Variance≅2.95c)If the standard deviation for the number of children per party is 1.72, then the variancewill be 2.95