Question

Please help me with these problems right now.

Answer 1

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.

Answer 2

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.