Please help me with these problems right now.

From normal probability distribution,
μ = mean = 3550
σ = standard deviation = 870
#47
Calculate z=(X- μ )/σ for lower and upper limits,
z1=(1810-3550)/870 = -2
z2=(4420-3550)/870 = 1
Probability of weights falling between 1810 and 4420
=P(z1<z<z2)
=P(-2<z<1)
=P(z<1)- P(z<1)
look up left-tale tables for normal distribution to get
P(z<1)=0.8413447
P(z<-2)=0.02275013
So
=P(z<1)- P(z<1)
=0.8413447-0.02275013
=0.81859
=> P(1810 < W < 4420) = 81.86%
#48
z=(4000-3550)/870
=0.5172414
P(W>4000)
=P(z>0.5172414)
=1-P(z<0.5172414)
Again, look up probability tables for normal distribution for P(z<0.5172414)
P(z<0.5172414)
=0.6975062
so
1-P(z<0.5172414)
=1-0.6975062
=0.3024938
=> P(W>4000) = 30.25%
Note: If a table of 4-digit accuracy is used, numbers may come out slightly differently.
47. The lower limit of your range of interest has a z-score of
... z = (x -μ)/σ = (1810 -3550)/870 = -2
The empirical rule tells you 95% of the distribution lies within ±2σ of the mean, so 5% lies in the two tails. Then 2.5% lies in the lower tail of the distribution beyond -2σ.
The upper limit of your range of interest has a z-score of
... z = (4420 -3550)/870 = 1
The empirical rule tells you 68% of the distribution lies within ±1σ of the mean, so 32% lies in the two tails. Then 16% lies in the upper tail beyond +1σ.
The sum of the lower tail and the upper tail beyond your range of interest is
... 2.5% + 16% = 18.5%
Thus the percentage of cars in the weight range of interest is
... 100% - 18.5% = 81.5% . . . . . selection d
48. My calculator says P(x>4000) = 0.30249383... An on-line calculator (shown below) shows the result as 30.25% (in agreement with my calculator). Since one of your selections appears to be 30.42%, we suspect a transposition error somewere along the line. Your best choice is selection b.