Answer: 0.628
Explanation:
Probability of U.S. smartphone owners have made an effort to limit their phone use in the past : p = 0.4702
Sample size : n= 54
Mean :
![\mu =(54)(0.4702)=25.39](https://img.qammunity.org/2021/formulas/mathematics/college/lnn4ka8doy5wdmrlungnslyzrzoxtmsgtu.png)
Standard deviation:
![\sigma=√((54)(0.4702))=√(25.39)\\\\\approx5.039](https://img.qammunity.org/2021/formulas/mathematics/college/v5rtu6en4i88iqvge4l6k5azesqk3q9t1b.png)
The probability that between 22 and 29 (inclusively) will have attempted to limit their cell phone use in the past will be :
![P(22\leq x\leq29)=P(21<x<30)=P((21-25.39)/(5.039)<(X-\mu)/(\sigma)<(30-25.39)/(5.039))\\\\=P(-0.8712<z<0.914864)\ \ \ \ [z=(X-\mu)/(\sigma)]\\\\=P(z<0.914864)-P(z<-0.8712)\\\\=P(z<0.914864)-(1-P(z<0.8712))\\\\=0.8198685-(1-0.8081775)\ \ \ [\text{ By p-value table}]\\\\=0.628046](https://img.qammunity.org/2021/formulas/mathematics/college/1m0a40670kgt6kbn6v2cv87zvg8qqknkwg.png)
Hence, the required probability = 0.628